In the previous notes, we defined the Lebesgue measure of a Lebesgue measurable set
, and set out the basic properties of this measure. In this set of notes, we use Lebesgue measure to define the Lebesgue integral
of functions . Just as not every set can be measured by Lebesgue measure, not every function can be integrated by the Lebesgue integral; the function will need to be Lebesgue measurable. Furthermore, the function will either need to be unsigned (taking values on
), or absolutely integrable.
To motivate the Lebesgue integral, let us first briefly review two simpler integration concepts. The first is that of an infinite summation
of a sequence of numbers , which can be viewed as a discrete analogue of the Lebesgue integral. Actually, there are two overlapping, but different, notions of summation that we wish to recall here. The first is that of the unsigned infinite sum, when the
lie in the extended non-negative real axis
. In this case, the infinite sum can be defined as the limit of the partial sums
or equivalently as a supremum of arbitrary finite partial sums:
The unsigned infinite sum always exists, but its value may be infinite, even when each term is individually finite (consider e.g.
).
The second notion of a summation is the absolutely summable infinite sum, in which the lie in the complex plane
and obey the absolute summability condition
where the left-hand side is of course an unsigned infinite sum. When this occurs, one can show that the partial sums converge to a limit, and we can then define the infinite sum by the same formula (1) as in the unsigned case, though now the sum takes values in
rather than
. The absolute summability condition confers a number of useful properties that are not obeyed by sums that are merely conditionally convergent; most notably, the value of an absolutely convergent sum is unchanged if one rearranges the terms in the series in an arbitrary fashion. Note also that the absolutely summable infinite sums can be defined in terms of the unsigned infinite sums by taking advantage of the formulae
for complex absolutely summable , and
for real absolutely summable , where
and
are the (magnitudes of the) positive and negative parts of
.
In an analogous spirit, we will first define an unsigned Lebesgue integral of (measurable) unsigned functions
, and then use that to define the absolutely convergent Lebesgue integral
of absolutely integrable functions
. (In contrast to absolutely summable series, which cannot have any infinite terms, absolutely integrable functions will be allowed to occasionally become infinite. However, as we will see, this can only happen on a set of Lebesgue measure zero.)
To define the unsigned Lebesgue integral, we now turn to another more basic notion of integration, namely the Riemann integral of a Riemann integrable function
. Recall from the prologue that this integral is equal to the lower Darboux integral
(It is also equal to the upper Darboux integral; but much as the theory of Lebesgue measure is easiest to define by relying solely on outer measure and not on inner measure, the theory of the unsigned Lebesgue integral is easiest to define by relying solely on lower integrals rather than upper ones; the upper integral is somewhat problematic when dealing with “improper” integrals of functions that are unbounded or are supported on sets of infinite measure.) Compare this formula also with (2). The integral is a piecewise constant integral, formed by breaking up the piecewise constant functions
into finite linear combinations of indicator functions of intervals, and then measuring the length of each interval.
It turns out that virtually the same definition allows us to define a lower Lebesgue integral of any unsigned function
, simply by replacing intervals with the more general class of Lebesgue measurable sets (and thus replacing piecewise constant functions with the more general class of simple functions). If the function is Lebesgue measurable (a concept that we will define presently), then we refer to the lower Lebesgue integral simply as the Lebesgue integral. As we shall see, it obeys all the basic properties one expects of an integral, such as monotonicity and additivity; in subsequent notes we will also see that it behaves quite well with respect to limits, as we shall see by establishing the two basic convergence theorems of the unsigned Lebesgue integral, namely Fatou’s lemma and the monotone convergence theorem.
Once we have the theory of the unsigned Lebesgue integral, we will then be able to define the absolutely convergent Lebesgue integral, similarly to how the absolutely convergent infinite sum can be defined using the unsigned infinite sum. This integral also obeys all the basic properties one expects, such as linearity and compatibility with the more classical Riemann integral; in subsequent notes we will see that it also obeys a fundamentally important convergence theorem, the dominated convergence theorem. This convergence theorem makes the Lebesgue integral (and its abstract generalisations to other measure spaces than ) particularly suitable for analysis, as well as allied fields that rely heavily on limits of functions, such as PDE, probability, and ergodic theory.
Remark 1 This is not the only route to setting up the unsigned and absolutely convergent Lebesgue integrals. Stein-Shakarchi, for instance, proceeds slightly differently, beginning with the unsigned integral but then making an auxiliary stop at integration of functions that are bounded and are supported on a set of finite measure, before going to the absolutely convergent Lebesgue integral. Another approach (which will not be discussed here) is to take the metric completion of the Riemann integral with respect to the
metric.
The Lebesgue integral and Lebesgue measure can be viewed as completions of the Riemann integral and Jordan measure respectively. This means three things. Firstly, the Lebesgue theory extends the Riemann theory: every Jordan measurable set is Lebesgue measurable, and every Riemann integrable function is Lebesgue measurable, with the measures and integrals from the two theories being compatible. Conversely, the Lebesgue theory can be approximated by the Riemann theory; as we saw in the previous notes, every Lebesgue measurable set can be approximated (in various senses) by simpler sets, such as open sets or elementary sets, and in a similar fashion, Lebesgue measurable functions can be approximated by nicer functions, such as Riemann integrable or continuous functions. Finally, the Lebesgue theory is complete in various ways; we will formalise this properly only in the next quarter when we study spaces, but the convergence theorems mentioned above already hint at this completeness. A related fact, known as Egorov’s theorem, asserts that a pointwise converging sequence of functions can be approximated as a (locally) uniformly converging sequence of functions. The facts listed here manifestations of Littlewood’s three principles of real analysis, which capture much of the essence of the Lebesgue theory.
— 1. Integration of simple functions —
Much as the Riemann integral was set up by first using the integral for piecewise constant functions, the Lebesgue integral is set up using the integral for simple functions.
Definition 1 (Simple function) A (complex-valued) simple function
is a finite linear combination
of indicator functions
of Lebesgue measurable sets
for
, where
is a natural number and
are complex numbers. An unsigned simple function
, is defined similarly, but with the
taking values in
rather than
.
It is clear from construction that the space of complex-valued simple functions forms a complex vector space; also,
also closed under pointwise product
and complex conjugation
. In short,
is a commutative
-algebra. Meanwhile, the space
of unsigned simple functions is a
-module; it is closed under addition, and under scalar multiplication by elements in
.
In this definition, we did not require the to be disjoint. However, it is easy enough to arrange this, basically by exploiting Venn diagrams (or, to use fancier language, finite boolean algebras). Indeed, any
subsets
of
partition
into
disjoint sets, each of which is an intersection of
or the complement
for
(and in particular, is measurable). The (complex or unsigned) simple function is constant on each of these sets, and so can easily be decomposed as a linear combination of the indicator function of these sets. One easy consequence of this is that if
is a complex-valued simple function, then its absolute value
is an unsigned simple function.
It is geometrically intuitive that we should define the integral of an indicator function of a measurable set
to equal
:
Using this and applying the laws of integration formally, we are led to propose the following definition for the integral of an unsigned simple function:
Definition 2 (Integral of a unsigned simple function) If
is an unsigned simple function, the integral
is defined by the formula
thus
will take values in
.
However, one has to actually check that this definition is well-defined, in the sense that different representations
of a function as a finite unsigned combination of indicator functions of measurable sets will give the same value for the integral . This is the purpose of the following lemma:
Lemma 3 (Well-definedness of simple integral) Let
be natural numbers,
, and let
be Lebesgue measurable sets such that the identity
holds identically on
. Then one has
Proof: We again use a Venn diagram argument. The sets
partition
into
disjoint sets, each of which is an intersection of some of the
and their complements. We throw away any sets that are empty, leaving us with a partition of
into
non-empty disjoint sets
for some
. As the
are Lebesgue measurable, the
are too. By construction, each of the
arise as unions of some of the
, thus we can write
and
for all and
, and some subsets
. By finite additivity of Lebesgue measure, we thus have
and
Thus, our objective is now to show that
To obtain this, we fix and evaluate (4) at a point
in the non-empty set
. At such a point,
is equal to
, and similarly
is equal to
. From (4) we conclude that
Multiplying this by and then summing over all
we obtain (5).
We now make some important definitions that we will use repeatedly in the course:
Definition 4 (Almost everywhere and support) A property
of a point
is said to hold (Lebesgue) almost everywhere in
, or for (Lebesgue) almost every point
, if the set of
for which
fails has Lebesgue measure zero (i.e.
is true outside of a null set). We usually omit the prefix Lebesgue, and often abbreviate “almost everywhere” or “almost every” as a.e.
Two functions
into an arbitrary range
are said to agree almost everywhere if one has
for almost every
.
The support of a function
or
is defined to be the set
where
is non-zero.
Note that if holds for almost every
, and
implies
, then
holds for almost every
. Also, if
are an at most countable family of properties, each of which individually holds for almost every
, then they will simultaneously be true for almost every
, because the countable union of null sets is still a null set. Because of these properties, one can (as a rule of thumb) treat the almost universal quantifier “for almost every” as if it was the truly universal quantifier “for every”, as long as one is only concatenating at most countably many properties together, and as long as one never specialises the free variable
to a null set. Observe also that the property of agreeing almost everywhere is an equivalence relation, which we will refer to as almost everywhere equivalence.
In later notes we will also see the notion of the closed support of a function , defined as the closure of the support.
The following properties of the simple unsigned integral are easily obtained from the definitions:
Exercise 1 (Basic properties of the simple unsigned integral) Let
be simple unsigned functions.
- (Unsigned linearity) We have
and
for all
.
- (Finiteness) We have
if and only if
is finite almost everywhere, and its support has finite measure.
- (Vanishing) We have
if and only if
is zero almost everywhere.
- (Equivalence) If
and
agree almost everywhere, then
.
- (Monotonicity) If
for almost every
, then
.
- (Compatibility with Lebesgue measure) For any Lebesgue measurable
, one has
.
Furthermore, show that the simple unsigned integral
is the only map from the space
of unsigned simple functions to
that obeys all of the above properties.
We can now define an absolutely convergent counterpart to the simple unsigned integral. This integral will be superceded by the absolutely Lebesgue integral, but we give it here as motivation for that more general notion of integration.
Definition 5 (Absolutely convergent simple integral) A complex-valued simple function
is said to be absolutely integrable if
. If
is absolutely integrable, the integral
is defined for real signed
by the formula
where
and
(note that these are unsigned simple functions that are pointwise dominated by
and thus have finite integral), and for complex-valued
by the formula
(Strictly speaking, this is an abuse of notation as we have now defined the simple integral
three different times, for unsigned, real signed, and complex-valued simple functions, but one easily verifies that these three definitions agree with each other on their common domains of definition, so it is safe to use a single notation for all three.)
Note from the preceding exercise that a complex-valued simple function is absolutely integrable if and only if it has finite measure support (since finiteness almost everywhere is automatic). In particular, the space
of absolutely integrable simple functions is closed under addition and scalar multiplication by complex numbers, and is thus a complex vector space.
The properties of the unsigned simple integral then can be used to deduce analogous properties for the complex-valued integral:
Exercise 2 (Basic properties of the complex-valued simple integral) Let
be absolutely integrable simple functions.
- (*-linearity) We have
- (Equivalence) If
and
agree almost everywhere, then
.
- (Compatibility with Lebesgue measure) For any Lebesgue measurable
, one has
.
(Hints: Work out the real-valued counterpart of the linearity property first. To establish (6), treat the cases
separately. To deal with the additivity for real functions
, start with the identity
and rearrange the second inequality so that no subtraction appears.) Furthermore, show that the complex-valued simple integral
is the only map from the space
of absolutely integrable simple functions to
that obeys all of the above properties.
We now comment further on the fact that (simple) functions that agree almost everywhere, have the same integral. We can view this as an assertion that integration is a noise-tolerant operation: one can have “noise” or “errors” in a function on a null set, and this will not affect the final value of the integral. Indeed, once one has this noise tolerance, one can even integrate functions
that are not defined everywhere on
, but merely defined almost everywhere on
(i.e.
is defined on some set
where
is a null set), simply by extending
to all of
in some arbitrary fashion (e.g. by setting
equal to zero on
). This is extremely convenient for analysis, as there are many natural functions (e.g.
in one dimension, or
for various
in higher dimensions) that are only defined almost everywhere instead of everywhere (often due to “division by zero” problems when a denominator vanishes). While such functions cannot be evaulated at certain singular points, they can still be integrated (provided they obey some integrability condition, of course, such as absolute integrability), and so one can still perform a large portion of analysis on such functions.
In fact, in the subfield of analysis known as functional analysis, it is convenient to abstract the notion of an almost everywhere defined function somewhat, by replacing any such function with the equivalence class of almost everywhere defined functions that are equal to
almost everywhere. Such classes are then no longer functions in the standard set-theoretic sense (they do not map each point in the domain to a unique point in the range, since points in
have measure zero), but the properties of various function spaces improve when one does this (various semi-norms become norms, various topologies become Hausdorff, and so forth). See these 245B lecture notes for further discussion.
Remark 2 The “Lebesgue philosophy” that one is willing to lose control on sets of measure zero is a perspective that distinguishes Lebesgue-type analysis from other types of analysis, most notably that of descriptive set theory, which is also interested in studying subsets of
, but can give completely different structural classifications to a pair of sets that agree almost everywhere. This loss of control on null sets is the price one has to pay for gaining access to the powerful tool of the Lebesgue integral; if one needs to control a function at absolutely every point, and not just almost every point, then one often needs to use other tools than integration theory (unless one has some regularity on the function, such as continuity, that lets one pass from almost everywhere true statements to everywhere true statements).
— 2. Measurable functions —
Much as the piecewise constant integral can be completed to the Riemann integral, the unsigned simple integral can be completed to the unsigned Lebesgue integral, by extending the class of unsigned simple functions to the larger class of unsigned Lebesgue measurable functions. One of the shortest ways to define this class is as follows:
Definition 6 (Unsigned measurable function) An unsigned function
is unsigned Lebesgue measurable, or measurable for short, if it is the pointwise limit of unsigned simple functions, i.e. if there exists a sequence
of unsigned simple functions such that
for every
.
This particular definition is not always the most tractable. Fortunately, it has many equivalent forms:
Lemma 7 (Equivalent notions of measurability) Let
be an unsigned function. Then the following are equivalent:
is unsigned Lebesgue measurable.
is the pointwise limit of unsigned simple functions
(thus the limit
exists and is equal to
for all
).
is the pointwise almost everywhere limit of unsigned simple functions
(thus the limit
exists and is equal to
for almost every
).
is the supremum
of an increasing sequence
of unsigned simple functions
, each of which are bounded with finite measure support.
- For every
, the set
is Lebesgue measurable.
- For every
, the set
is Lebesgue measurable.
- For every
, the set
is Lebesgue measurable.
- For every
, the set
is Lebesgue measurable.
- For every interval
, the set
is Lebesgue measurable.
- For every (relatively) open set
, the set
is Lebesgue measurable.
- For every (relatively) closed set
, the set
is Lebesgue measurable.
Proof: (1.) and (2.) are equivalent by definition. (2.) clearly implies (3.). As every monotone sequence in converges, (4.) implies (2.). Now we show that (3.) implies (5.). If
is the pointwise almost everywhere limit of
, then for almost every
one has
This implies that, for any , the set
is equal to
outside of a set of measure zero; this set in turn is equal to
outside of a set of measure zero. But as each is an unsigned simple function, the sets
are Lebesgue measurable. Since countable unions or countable intersections of Lebesgue measurable sets are Lebesgue measurable, and modifying a Lebesgue measurable set on a null set produces another Lebesgue measurable set, we obtain (5.).
To obtain the equivalence of (5.) and (6.), observe that
for and
, where
are the non-negative rationals. The claim then easily follows from the countable nature of
(treating the extreme cases
separately if necessary). A similar argument lets one deduce (5.) or (6.) from (9.).
The equivalence of (5.), (6.) with (7.), (8.) comes from the observation that is the complement of
, and
is the complement of
. A similar argument shows that (10.) and (11.) are equivalent.
By expressing an interval as the intersection of two half-intervals, we see that (9.) follows from (5.)-(8.), and so all of (5.)-(9.) are now shown to be equivalent.
Clearly (10.) implies (7.), and hence (5.)-(9.). Conversely, because every open set in is the union of countably many open intervals in
, (9.) implies (10.).
The only remaining task is to show that (5.)-(11.) implies (4.). Let obey (5.)-(11.). For each positive integer
, we let
be defined to be the largest integer multiple of
that is less than or equal to
when
, with
for
. From construction it is easy to see that the
are increasing and have
as their supremum. Furthermore, each
takes on only finitely many values, and for each non-zero value
it attains, the set
takes the form
for some interval or ray
, and is thus measurable. As a consequence,
is a simple function, and by construction it is bounded and has finite measure support. The claim follows.
With these equivalent formulations, we can now generate plenty of measurable functions:
Exercise 3
- Show that every continuous function
is measurable.
- Show that every unsigned simple function is measurable.
- Show that the supremum, infimum, limit superior, or limit inferior of unsigned measurable functions is unsigned measurable.
- Show that an unsigned function that is equal almost everywhere to an unsigned measurable function, is itself measurable.
- Show that if a sequence
of unsigned measurable functions converges pointwise almost everywhere to an unsigned limit
, then
is also measurable.
- If
is measurable and
is continuous, show that
is measurable.
- If
are unsigned measurable functions, show that
and
are measurable.
In view of part (4.) of the above exercise, one can define the concept of measurability for an unsigned function that is only defined almost everywhere on , rather than everywhere on
, by extending that function arbitrarily to the null set where it is currently undefined.
Exercise 4 Let
. Show that
is a bounded unsigned measurable function if and only if
is the uniform limit of bounded simple functions.
Exercise 5 Show that an unsigned function
is a simple function if and only if it is measurable and takes on at most finitely many values.
Exercise 6 Let
be an unsigned measurable function. Show that the region
is a measurable subset of
. (There is a converse to this statement, but we will wait until later notes to prove it, once we have the Fubini-Tonelli theorem available to us.)
Remark 3 Lemma 7 tells us that if
is measurable, then
is Lebesgue measurable for many classes of sets
. However, we caution that it is not necessarily the case that
is Lebesgue measurable if
is Lebesgue measurable. To see this, we let
be the Cantor set
and let
be the function defined by setting
whenever
is not a terminating binary decimal, and so has a unique binary expansion
for some
, and
otherwise. We thus see that
takes values in
, and is bijective on the set
of non-terminating decimals in
. Using Lemma 7, it is not difficult to show that
is measurable. On the other hand, by modifying the construction from the previous notes, we can find a subset
of
which is non-measurable. If we set
, then
is a subset of the null set
and is thus itself a null set; but
is non-measurable, and so the inverse image of a Lebesgue measurable set by a measurable function need not remain Lebesgue measurable.
However, in later notes we will see that it is still true that
is Lebesgue measurable if
has a slightly stronger measurability property than Lebesgue measurability, namely Borel measurability.
Now we can define the concept of a complex-valued measurable function. As discussed earlier, it will be convenient to allow for such functions to only be defined almost everywhere, rather than everywhere, to allow for the possibility that the function becomes singular or otherwise undefined on a null set.
Definition 8 (Complex measurability) An almost everywhere defined complex-valued function
is Lebesgue measurable, or measurable for short, if it is the pointwise almost everywhere limit of complex-valued simple functions.
As before, there are several equivalent definitions:
Exercise 7 Let
be an almost everywhere defined complex-valued function. Then the following are equivalent:
is measurable.
is the pointwise almost everywhere limit of complex-valued simple functions.
- The (magnitudes of the) positive and negative parts of
and
are unsigned measurable functions.
is Lebesgue measurable for every open set
.
is Lebesgue measurable for every closed set
.
From the above exercise, we see that the notion of complex-valued measurability and unsigned measurability are compatible when applied to a function that takes values in everywhere (or almost everywhere).
Exercise 8
- Show that every continuous function
is measurable.
- Show that a function
is simple if and only if it is measurable and takes on at most finitely many values.
- Show that a complex-valued function that is equal almost everywhere to an measurable function, is itself measurable.
- Show that if a sequence
of complex-valued measurable functions converges pointwise almost everywhere to an complex-valued limit
, then
is also measurable.
- If
is measurable and
is continuous, show that
is measurable.
- If
are measurable functions, show that
and
are measurable.
Exercise 9 Let
be a Riemann integrable function. Show that if one extends
to all of
by defining
for
, then
is measurable.
— 3. Unsigned Lebesgue integrals —
We are now ready to integrate unsigned measurable functions. We begin with the notion of the lower unsigned Lebesgue integral, which can be defined for arbitrary unsigned functions (not necessarily measurable):
Definition 9 (Lower unsigned Lebesgue integral) Let
be an unsigned function (not necessarily measurable). We define the lower unsigned Lebesgue integral
to be the quantity
where
ranges over all unsigned simple functions
that are pointwise bounded by
.
One can also define the upper unsigned Lebesgue integral
but we will use this integral much more rarely. Note that both integrals take values in
, and that the upper Lebesgue integral is always at least as large as the lower Lebesgue integral.
In the definition of the lower unsigned Lebesgue integral, is required to be bounded by
pointwise everywhere, but it is easy to see that one could also require
to just be bounded by
pointwise almost everywhere without affecting the value of the integral, since the simple integral is not affected by modifications on sets of measure zero.
The following properties of the lower Lebesgue integral are easy to establish:
Exercise 10 (Basic properties of the lower Lebesgue integral) Let
be unsigned functions (not necessarily measurable).
- (Compatibility with the simple integral) If
is simple, then
.
- (Monotonicity) If
pointwise almost everywhere, then
and
.
- (Homogeneity) If
, then
.
- (Equivalence) If
agree almost everywhere, then
and
.
- (Superadditivity)
.
- (Subadditivity of upper integral)
- (Divisibility) For any measurable set
, one has
.
- (Horizontal truncation) As
,
converges to
.
- (Vertical truncation) As
,
converges to
. Hint: From Exercise 11 of Notes 1, we have
for any measurable set
.
- (Reflection) If
is a simple function that is bounded with finite measure support (i.e. it is absolutely integrable), then
.
Do the horizontal and vertical truncation properties hold if the lower Lebesgue integral is replaced with the upper Lebesgue integral?
Now we restrict attention to measurable functions.
Definition 10 (Unsigned Lebesgue integral) If
is measurable, we define the unsigned Lebesgue integral
of
to equal the lower unsigned Lebesgue integral
. (For non-measurable functions, we leave the unsigned Lebesgue integral undefined.)
One nice feature of measurable functions is that the lower and upper Lebesgue integrals can match, if one also assumes some boundedness:
Exercise 11 Let
be measurable, bounded, and vanishing outside of a set of finite measure. Show that the lower and upper Lebesgue integrals of
agree. (Hint: use Exercise 4.) There is a converse to this statement, but we will defer it to later notes. What happens if
is allowed to be unbounded, or is not supported inside a set of finite measure?
This gives an important corollary:
Corollary 11 (Finite additivity of the Lebesgue integral) Let
be measurable. Then
.
Proof: From the horizontal truncation property and a limiting argument, we may assume that are bounded. From the vertical truncation property and another limiting argument, we may assume that
are supported inside a bounded set. From Exercise 11, we now see that the lower and upper Lebesgue integrals of
,
, and
agree. The claim now follows by combining the superadditivity of the lower Lebesgue integral with the subadditivity of the upper Lebesgue integral.
In later notes we will improve this finite additivity property for the unsigned Lebesgue integral further, to countable additivity; this property is also known as the monotone convergence theorem.
Exercise 12 (Upper Lebesgue integral and outer Lebesgue measure) Show that for any set
,
. Conclude that the upper and lower Lebesgue integrals are not necessarily additive if no measurability hypotheses are assumed.
Exercise 13 (Area interpretation of integral) If
is measurable, show that
is equal to the
-dimensional Lebesgue measure of the region
. (This can be used as an alternate, and more geometrically intuitive, definition of the unsigned Lebesgue integral; it is a more convenient formulation for establishing the basic convergence theorems, but not quite as convenient for establishing basic properties such as additivity.) (Hint: use Exercise 22 from Notes 1.)
Exercise 14 Show that the Lebesgue integral
is the only map from measurable unsigned functions
to
that obeys the following properties for measurable
:
- (Compatibility with the simple integral) If
is simple, then
.
- (Finite additivity)
.
- (Horizontal truncation) As
,
converges to
.
- (Vertical truncation) As
,
converges to
.
Exercise 15 (Translation invariance) Let
be measurable. Show that
for any
.
Exercise 16 (Linear change of variables) Let
be measurable, and let
be an invertible linear transformation. Show that
, or equivalently
.
Exercise 17 (Compatibility with the Riemann integral) Let
be Riemann integrable. If we extend
to
by declaring
to equal zero outside of
, show that
.
We record a basic inequality, known as Markov’s inequality, that asserts that the Lebesgue integral of an unsigned measurable function controls how often that function can be large:
Lemma 12 (Markov’s inequality) Let
be measurable. Then for any
, one has
Proof: We have the trivial pointwise inequality
From the definition of the lower Lebesgue integral, we conclude that
and the claim follows.
By sending to infinity or to zero, we obtain the following important corollary:
Exercise 18 Let
be measurable.
- Show that if
, then
is finite almost everywhere. Give a counterexample to show that the converse statement is false.
- Show that
if and only if
is zero almost everywhere.
Remark 4 The use of the integral
to control the distribution of
is known as the first moment method. One can also control this distribution using higher moments such as
for various values of
, or exponential moments such as
or the Fourier moments
for various values of
; such moment methods are fundamental to probability theory.
— 4. Absolute integrability —
Having set out the theory of the unsigned Lebesgue integral, we can now define the absolutely convergent Lebesgue integral.
Definition 13 (Absolute integrability) An almost everywhere defined measurable function
is said to be absolutely integrable if the unsigned integral
is finite. We refer to this quantity
as the
norm of
, and use
or
to denote the space of absolutely integrable functions. If
is real-valued and absolutely integrable, we define the Lebesgue integral
by the formula
where
,
are the magnitudes of the positive and negative components of
(note that the two unsigned integrals on the right-hand side are finite, as
are pointwise dominated by
). If
is complex-valued and absolutely integrable, we define the Lebesgue integral
by the formula
where the two integrals on the right are interpreted as real-valued absolutely integrable Lebesgue integrals. It is easy to see that the unsigned, real-valued, and complex-valued Lebesgue integrals defined in this manner are compatible on their common domains of definition.
Note from construction that the absolutely integrable Lebesgue integral extends the absolutely integrable simple integral, which is now redundant and will not be needed any further in the sequel.
Remark 5 One can attempt to define integrals for non-absolutely-integrable functions, analogous to the improper integrals
or the principal value integrals
one sees in the classical one-dimensional Riemannian theory. While one can certainly generate any number of such extensions of the Lebesgue integral concept, such extensions tend to be poorly behaved with respect to various important operations, such as change of variables or exchanging limits and integrals, so it is usually not worthwhile to try to set up a systematic theory for such non-absolutely-integrable integrals that is anywhere near as complete as the absolutely integrable theory, and instead deal with such exotic integrals on an ad hoc basis.
From the pointwise triangle inequality , we conclude the
triangle inequality
for any almost everywhere defined measurable . It is also easy to see that
for any complex number . As such, we see that
is a complex vector space. (The
norm is then a seminorm on this space, but we will not need to discuss norms and seminorms in detail until 245B.) From Exercise 18 we make the important observation that a function
has zero
norm,
, if and only if
is zero almost everywhere.
Given two functions , we can define the
distance
between them by the formula
Thanks to (8), this distance obeys almost all the axioms of a metric on , with one exception: it is possible for two different functions
to have a zero
distance, if they agree almost everywhere. As such,
is only a semi-metric (also known as a pseudo-metric) rather than a metric. However, if one adopts the convention that any two functions that agree almost everywhere are considered equivalent (or more formally, one works in the quotient space of
by the equivalence relation of almost everywhere agreement, which by abuse of notation is also denoted
), then one recovers a genuine metric. (Later on, we will establish the important fact that this metric makes the (quotient space)
a complete metric space, a fact known as the
Riesz-Fischer theorem; this completeness is one of the main reasons we spend so much effort setting up Lebesgue integration theory in the first place.)
The linearity properties of the unsigned integral induce analogous linearity properties of the absolutely convergent Lebesgue integral:
Exercise 19 (Integration is linear) Show that integration
is a (complex) linear operation from
to
. In other words, show that
and
for all absolutely integrable
and complex numbers
. Also establish the identity
which makes integration not just a linear operation, but a *-linear operation.
Exercise 20 Show that Exercises 15, 16, and 17 also hold for complex-valued, absolutely integrable functions rather than for unsigned measurable functions.
Exercise 21 (Absolute summability is a special case of absolute integrability) Let
be a doubly infinite sequence of complex numbers, and let
be the function
where
is the greatest integer less than
. Show that
is absolutely integrable if and only if the series
is absolutely convergent, in which case one has
.
We can localise the absolutely convergent integral to any measurable subset of
. Indeed, if
is a function, we say that
is measurable (resp. absolutely integrable) if its extension
is measurable (resp. absolutely integrable), where
is defined to equal
when
and zero otherwise, and then we define
. Thus, for instance, the absolutely integrable analogue of Exercise 17 tells us that
for any Riemann-integrable .
Exercise 22 If
are disjoint measurable subsets of
, and
is absolutely integrable, show that
and
We will study the properties of the absolutely convergent Lebesgue integral in more detail in later notes, as a special case of the more general Lebesgue integration theory on abstract measure spaces. For now, we record one very basic inequality:
Lemma 14 (Triangle inequality) Let
. Then
Proof: If is real-valued, then
and the claim is obvious from (7). When
is complex-valued, one cannot argue quite so simply; a naive mimicking of the real-valued argument would lose a factor of
, giving the inferior bound
To do better, we exploit the phase rotation invariance properties of the absolute value operation and of the integral, as follows. Note that for any complex number , one can write
as
for some real
. In particular, we have
for some real . Taking real parts of both sides, we obtain
Since , we obtain the claim.
— 5. Littlewood’s three principles —
Littlewood’s three principles are informal heuristics that convey much of the basic intuition behind the measure theory of Lebesgue. Briefly, the three principles are as follows:
- Every (measurable) set is nearly a finite sum of intervals;
- Every (absolutely integrable) function is nearly continuous; and
- Every (pointwise) convergent sequence of functions is nearly uniformly convergent.
Various manifestations of the first principle were given in the previous set of notes (and specifically, in Exercises 7 and 14 of those notes). We now discuss the second principle. Define a step function to be a finite linear combination of indicator functions of boxes.
Theorem 15 (Approximation of
functions) Let
and
.
- There exists an absolutely integrable simple function
such that
.
- There exists a step function
such that
.
- There exists a continuous, compactly supported
such that
.
To put things another way, the absolutely integrable simple functions, the step functions, and the continuous, compactly supported functions are all dense subsets of with respect to the
(semi-)metric. Much later in the course (in 245C), we will see that a similar statement holds if one replaces continuous, compactly supported functions with smooth, compactly supported functions, also known as test functions; this is an important fact for the theory of distributions.
Proof: We begin with part (1.). When is unsigned, we see from the definition of the lower Lebesgue integral that there exists an unsigned simple function
such that
(so, in particular,
is absolutely integrable) and
which by linearity implies that . This gives (1.) when
is unsigned. The case when
is real-valued then follows by splitting
into positive and negative parts (and adjusting
as necessary), and the case when
is complex-valued then follows by splitting
into real and imaginary parts (and adjusting
yet again).
To establish part (2.), we see from (1.) and the triangle inequality in that it suffices to show this when
is an absolutely integrable simple function. By linearity (and more applications of the triangle inequality), it then suffices to show this when
is the indicator function of a measurable set
of finite measure. But then, by Exercise 14 of Notes 1, such a set can be approximated (up to an error of measure at most
) by an elementary set, and the claim follows.
To establish part (3.), we see from (2.) and the argument from the preceding paragraph that it suffices to show this when is the indicator function of a box. But one can then establish the claim by direct construction. Indeed, if one makes a slightly larger box
that contains the closure of
in its interior, but has a volume at most
more than that of
, then one can directly construct a piecewise linear continuous function
supported on
that equals
on
(e.g. one can set
for some sufficiently large
; one may also invoke Urysohn’s lemma, which we wil cover in 245B). It is then clear from construction that
as required.
This is not the only way to make Littlewood’s second principle manifest; we return to this point shortly. For now, we turn to Littlewood’s third principle. We recall three basic ways in which a sequence of functions can converge to a limit
:
- (Pointwise convergence)
for every
.
- (Pointwise almost everywhere convergence)
for almost every
.
- (Uniform convergence) For every
, there exists
such that
for all
and all
.
Uniform convergence implies pointwise convergence, which in turn implies pointwise almost everywhere convergence.
We now add a fourth mode of convergence, that is weaker than uniform convergence but stronger than pointwise convergence:
Definition 16 (Locally uniform convergence) A sequence of functions
converges locally uniformly to a limit
if, for every bounded subset
of
,
converges uniformly to
on
. In other words, for every bounded
and every
, there exists
such that
for all
and
.
Remark 6 At least as far as
is concerned, an equivalent definition of local uniform convergence is:
converges locally uniformly to
if, for every point
, there exists an open neighbourhood
of
such that
converges uniformly to
on
. The equivalence of the two definitions is immediate from the Heine-Borel theorem. More generally, the adverb “locally” in mathematics is usually used in this fashion; a propery
is said to hold locally on some domain
if, for every point
in that domain, there is an open neighbourhood of
in
on which
holds.
One should caution, though, that on domains on which the Heine-Borel theorem does not hold, the bounded-set notion of local uniform convergence is not equivalent to the open-set notion of local uniform convergence (though, for locally compact spaces, one can recover equivalence if one replaces “bounded” by “compact”).
Example 1 The functions
on
for
converge locally uniformly (and hence pointwise) to zero on
, but do not converge uniformly.
Example 2 The partial sums
of the Taylor series
converges to
locally uniformly (and hence pointwise) on
, but not uniformly.
Example 3 The functions
for
(with the convention that
) converge pointwise everywhere to zero, but do not converge locally uniformly.
From the preceding example, we see that pointwise convergence (either everywhere or almost everywhere) is a weaker concept than local uniform convergence. Nevertheless, a remarkable theorem of Egorov, which demonstrates Littlewood’s third principle, asserts that one can recover local uniform convergence as long as one is willing to delete a set of small measure:
Theorem 17 (Egorov’s theorem) Let
be a sequence of measurable functions that converge pointwise almost everywhere to another function
, and let
. Then there exists a Lebesgue measurable set
of measure at most
, such that
converges locally uniformly to
outside of
.
Note that Example 3 demonstrates that the exceptional set in Egorov’s theorem cannot be taken to have zero measure, at least if one uses the bounded-set definition of local uniform convergence from Definition 16. (If one instead takes the “open neighbourhood” definition, then the sequence in Example 3 does converge locally uniformly on
in the open neighbourhood sense, even if it does not do so in the bounded-set sense. On a domain such as
, bounded-set locally uniform convergence implies open-neighbourhood locally uniform convergence, but not conversely, so for the purposes of applying Egorov’s theorem, the distinction is not too important since one has local uniform convergence in both senses.)
Proof: By modifying and
on a set of measure zero (that can be absorbed into
at the end of the argument) we may assume that
converges pointwise everywhere to
, thus for every
and
there exists
such that
for all
. We can rewrite this fact set-theoretically as
for each , where
It is clear that the are Lebesgue measurable, and are decreasing in
. Applying downward monotone convergence (Exercise 9 of Notes 1) we conclude that, for any radius
, one has
(The restriction to the ball is necessary, because the downward monotone convergence property only works when the sets involved have finite measure.) In particular, for any
, we can find
such that
for all .
Now let . Then
is Lebesgue measurable, and by countable subadditivity,
. By construction, we have
whenever ,
,
, and
. In particular, we see for any ball
with an integer radius,
converges uniformly to
on
. Since every bounded set is contained in such a ball, the claim follows.
Remark 7 Unfortunately, one cannot in general upgrade local uniform convergence to uniform convergence in Egorov’s theorem. A basic example here is the moving bump example
on
. This sequence converges pointwise (and locally uniformly) to the zero function
. However, for any
and any
, we have
on a set of measure
, namely on the interval
. Thus, if one wanted
to converge uniformly to
outside of a set
, then that set
has to contain a set of measure
. In fact, it must contain the intervals
for all sufficiently large
and must therefore have infinite measure.
However, if all the
and
were supported on a fixed set
of finite measure (e.g. on a ball
), then the above “escape to horizontal infinity” cannot occur, it is easy to see from the above argument that one can recover uniform convergence (and not just locally uniform convergence) outside of a set of arbitrarily small measure.
We now use Theorem 15 to give another version of Littlewood’s second principle, known as Lusin’s theorem:
Theorem 18 (Lusin’s theorem) Let
be absolutely integrable, and let
. Then there exists a Lebesgue measurable set
of measure at most
such that the restriction of
to the complementary set
is continuous on that set.
Caution: this theorem does not imply that the unrestricted function is continuous on
. For instance, the absolutely integrable function
is nowhere continuous, so is certainly not continuous on
for any
of finite measure; but on the other hand, if one deletes the measure zero set
from the reals, then the restriction of
to
is identically zero and thus continuous.
Proof: By Theorem 15, for any one can find a continuous, compactly supported function
such that
(say). By Markov’s inequality, that implies that
for all
outside of a Lebesgue measurable set
of measure at most
. Letting
, we conclude that
is Lebesgue measurable with measure at most
, and
converges uniformly to
outside of
. But the uniform limit of continuous functions is continuous. We conclude that the restriction
to
is continuous, as required.
Exercise 23 Show that the hypothesis that
is absolutely integrable in Lusin’s theorem can be relaxed to being locally absolutely integrable (i.e. absolutely integrable on every bounded set), and then relaxed further to that of being measurable (but still finite everywhere or almost everywhere). (To achieve the latter goal, one can replace
locally with a horizontal truncation
; alternatively, one can replace
with a bounded variant, such as
.)
Exercise 24 Show that a function
is measurable if and only if it is the pointwise almost everywhere limit of continuous functions
. (Hint: if
is measurable and
, show that there exists a continuous function
for which the set
has measure at most
. You may find Exercise 25 below to be useful for this.) Use this (and Egorov’s theorem) to give an alternate proof of Lusin’s theorem for arbitrary measurable functions.
Remark 8 This is a trivial but important remark: when dealing with unsigned measurable functions such as
, then Lusin’s theorem does not apply directly because
could be infinite on a set of positive measure, which is clearly in contradiction with the conclusion of Lusin’s theorem (unless one allows the continuous function to also take values in the extended non-negative reals
with the extended topology). However, if one knows already that
is almost everywhere finite (which is for instance the case when
is absolutely integrable), then Lusin’s theorem applies (since one can simply zero out
on the null set where it is infinite, and add that null set to the exceptional set of Lusin’s theorem).
Remark 9 By combining Lusin’s theorem with inner regularity (Exercise 13 from Notes 1) and the Tietze extension theorem (which we will cover in 245B), one can conclude that every measurable function
agrees (outside of a set of arbitrarily small measure) with a continuous function
.
Exercise 25 (Littlewood-like principles) The following facts are not, strictly speaking, instances of any of Littlewood’s three principles, but are in a similar spirit.
- (Absolutely integrable functions almost have bounded support) Let
be an absolutely integrable function, and let
. Show that there exists a ball
outside of which
has an
norm of at most
, or in other words that
.
- (Measurable functions are almost locally bounded) Let
be a measurable function supported on a set of finite measure, and let
. Show that there exists a measurable set
of measure at most
outside of which
is locally bounded, or in other words that for every
there exists
such that
for all
.
As with Remark 8, it is important in the second part of the exercise that
is known to be finite everywhere (or at least almost everywhere); the result would of course fail if
was, say, unsigned but took the value
on a set of positive measure.
110 comments
Comments feed for this article
19 September, 2010 at 7:54 pm
David Roberts
typo: “but much as the theory of Lebesgue measure is easiest to define by relying solely on outer measure and noton inner measure…”
[Corrected, thanks – T.]
19 September, 2010 at 7:59 pm
David Roberts
And exercises 1 and 2 (and to a lesser extent definition 5) have not typeset properly: the equations break the bounding box and are cut off.
[Corrected, thanks – T.]
20 September, 2010 at 12:07 am
Bo Jacoby
“$latex{c_n^+ := \max(c_n,0)}$ and $latex{c_n^- := \max(-c_n,0)}$ are the positive and negative parts of $latex{c_n}$” means “$latex{c_n^+ := \max(c_n,0)}$ and $latex{-c_n^- := \min(c_n,0)}$ are the positive and negative parts of {c_n}”. ($latex{c_n^-}$ is neither negative nor part of $Latex{c_n}$).
[Reworded slightly – T.]
20 September, 2010 at 4:42 am
Thomas
The end of the proof of Theorem 15 appears to have gone awry. [Corrected, thanks – T.]
20 September, 2010 at 6:47 pm
Anonymous
in definition 1, is
or
?
thanks
20 September, 2010 at 6:59 pm
Terence Tao
It doesn’t matter much for this definition (either way, 0 will be a simple function).
21 September, 2010 at 3:42 am
Real Analysis (=Measure Theory) by Terence Tao « UGroh's Weblog
[…] Prolog: Grundfragen und Jordansche Maß Lecture 1: Das Lebesguesche Maß. Lecture 2: Das Lebesguesche Integral […]
21 September, 2010 at 12:30 pm
Mark Schwarzmann
Another possible definition of the unsigned Lebesgue integral arises from its interpretation as the area (or n-dimensional volume, in general) under the graph of a function. Formally, if
is a non-negative measurable function, one may define the integral of
on
as the
-dimensional Lebesgue measure of the set
, where $x \in \mathbb{R}^n$ and $y \in \mathbb{R}$. For this definition, the basic limit theorems of the Lebesgue integral (the monotone convergence theorem and Fatou's lemma) follow readily from the corresponding theorems for sets, though other properties of the integral are admittedly more difficult to establish.
[A comment added to Exercise 13 to reflect this – T.]
22 September, 2010 at 7:01 am
Laurent
Very nice series of posts, thank you!
should be defined to be the complement of what it is defined to be.
In the proof of Egorov’s theorem, it seems that
[Corrected, thanks – T.]
25 September, 2010 at 10:59 pm
245A, Notes 3: Integration on abstract measure spaces, and the convergence theorems « What’s new
[…] now define the unsigned integral, similarly to what was done for the unsigned Lebesgue integral in Notes 2: Definition 11 Let be a measure space, and let be measurable. Then we define the unsigned […]
6 October, 2010 at 10:41 pm
Anonymous
Dear Prof. Tao
Thanks for the nice posts!
I have a question regarding the definition of the simple function: It is defined to take finite many values (i=1,…k in Def. 1). I can see it is sufficient for the theory (e.g., the lower Lebesgue integral is taken to be the supremum of the integral of simple function). But would it be problematic if we allow k go to infinity in the Definition 1?
Thanks!
11 October, 2010 at 7:58 am
Terence Tao
Well, I suppose one could consider functions that take countably many values; given that the measure is countably additive, one would imagine that much of the theory for finitely-valued functions would carry over to countably-valued functions. But simple functions are in any event just a stepping stone to the integration of arbitrary unsigned measurable functions, which can take uncountably many values, so it’s not clear what advantage would be gained by making simple functions less simple.
9 October, 2010 at 7:44 am
chandrasekhar
Respected Sir,
Instead of using the ” \rightarrow” symbol for getting
just use “\to” for getting the same symbol, thereby saving time!
10 October, 2010 at 11:12 pm
Brent Nelson
In exercise 19, one of the complex constants ‘c’ should be moved inside the integral.
[Corrected, thanks – T.]
12 October, 2010 at 9:53 pm
Brent Nelson
In part (5) of exercise 7, the U should be changed to a K. In definition 10, you say that the unsigned Lebesgue integral is equal to the lower unsigned Lebesgue integral, but you use the notation for the upper unsigned Lebesgue integral. [Corrected, thanks – T.]
16 October, 2010 at 8:29 pm
245A, Notes 5: Differentiation theorems « What’s new
[…] integrable, and let be arbitrary. Applying Littlewood’s second principle (Theorem 15 from Notes 2) to the absolutely integrable function , we can find a continuous, compactly supported function […]
24 October, 2010 at 1:21 am
Anonymous
Hi Prof. Tao,
I have a question of Exercise 4. Why f is a bounded unsigned measurable function if it is uniform limit of simple function.
What if we take f_n equals infinity, and it is a simple function and uniformly convergent to f which also equals infinity.
[Oops, I had omitted the requirement that the simple functions must also be bounded. Fixed now – T.]
24 October, 2010 at 10:55 pm
Mark Schwarzmann
Another fact in the spirit of Littlewood’s three principles is that every absolutely integrable function nearly vanishes at infinity (that is, outside a set of arbitrarily small measure).
[A good point! I’ve added a final exercise to the notes to draw attention to this (and also to the related fact that a function that is finite a.e. is almost locally bounded.]
25 October, 2010 at 7:19 pm
Hossein Naderi
Dear Professor Tao,
I had a quick question on excersice 4 (The more difficult direction), why would boundedness of f as a measureable function cause uniform instead of pointwise convergence of simple functions?
Thank you in advance,
Hossein
25 October, 2010 at 8:19 pm
Terence Tao
To approximate arbitrary measurable functions by (bounded) simple functions, one has to perform two operations: vertical truncation and discretisation. (Horizontal truncation is not necessary because we are not requiring the simple functions to have finite measure support.) The latter would lead to uniform convergence, but the former does not, which is why one only has pointwise convergence in the arbitrary case. But in the bounded case, one does not need to perform vertical truncation, only discretisation, and this recovers the uniform convergence.
25 October, 2010 at 8:46 pm
Hossein Naderi
Thank you Professor Tao,
I got it. Really appreciate your help.
Hossein
17 November, 2010 at 9:56 am
Qiang
Dear Prof. Tao
In Remark 3, the second sentence should be … if
is NOT Lebesgue measurable… [Actually, I think the sentence is correct as it stands – T.]
1 December, 2010 at 3:21 pm
Anonymous
Prof. Tao,
instead of
? In many of the proofs of your notes, you truncate the simple functions because it can be unbounded by the definition. What if you just “define” that the unsigned simple functions are bounded? Then does your route to setting up the Lebesgue integrals still work?
In your route to setting up the unsigned and absolutely convergent Lebesgue integrals, could you explain what are the advantages for allowing the unsigned simple functions to be unbounded, say,
1 December, 2010 at 4:17 pm
Terence Tao
Yes, this is an alternate way to proceed. It moves some minor difficulties elsewhere; for instance, showing that two functions have the same integral if they differ on a set of measure zero becomes very slightly more difficult if simple functions are required to be finite-valued.
16 December, 2010 at 4:28 am
245A, Notes 2: The Lebesgue integral « mathTHÍCHinTOÁNmyHỌCbrain
[…] the partial sums converge to a limit, and we can then define the infinite sum by the same formula (1) as in the unsigned case, though now the sum takes values in rather than . The absolute summability […]
28 December, 2010 at 7:32 am
Anonymous
Dear professor Tao,
(actually I don’t know how you would like to be called)
I believe there is a mistake in the proof of lemma 7 (and I’m sorry if it is my mistake), more precisely in the statement (3) => (5).
If you take
to be constant and equal to
, clearly
pointwise. Take
. For a fixed
the set
is the whole
, so that the intersection of these sets is
, while the set of values for each
is empty, which shows the sets are in general different, isn’t it?
[Corrected, thanks – T.]
29 December, 2010 at 2:27 pm
Anonymous
Professor Tao,
In Lemma 7, how do you know the set
, for simple functions
, is Lebesgue measurable?
Thanks.
31 December, 2010 at 1:29 pm
Anonymous
Anonymous (above me),
by definition simple functions have finite image. Therefore, for any constant
,
is the union of a finite number of measurable sets, each of them being
for some constant
. And each of that sets is measurable by definition.
Dear professor Tao,
(I still don’t know how you would like to be called)
In exercise 10.3, of course remembering that
is not necessarily simple, I wonder if the tricky part to which you refered is not showing the equality for
, for the lower integral. It seems that the upper integral case is just about proving that a countable partition of a set with nonzero exterior measure yields some sets of nonzero exterior measure. Isn’t it?
If I am correct, could you please include some hints?
12 January, 2011 at 7:05 pm
Anonymous
Dear professor and readers,
actually I think there is a counter example to exercise 10.3, for
(which is of course a non-measurable function
. That is, there is
such that
.
Suppose we have found sets
such that
but each
contains no (measurable) set of positive measure. Take
. Any nonnegative simple function that is pointwise smaller than or equals to
which is
in a set of positive measure induces a subset of
with positive measure, which cannot exist. Thus
has a null integral, while
is infinite in
, and we get our example.
To find such sets, take
to be a Vitali set, i.e., a set which contains exactly one number in each coset
of the group
. If
is an enumeration of
, choose
.
A subset of
contains no set of positive measure if and only if its complement is not contained in any set of measure smaller than 1. This is the same as saying that it has outer measure 1, by outer regularity (each measurable set is contained in open sets which approximate its measure arbitrarily much).
Thus, we ought to prove that
has outer measure 1. Here there is a result similar to Lebesgue differentiation theorem (see notes 5, and more precisely exercise 24; but the needed result is here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=381857). Given
and set
of positive outer measure, there is an interval
such that the density
is
. By
-subadditivity we may suppose
bounded, so that we can cover it by open intervals whose total measure is
, and the result is not hard.
We already know that
has positive outer measure (exercise 25, notes 1), so that there are intervals
where
has a high density. Because
is dense in
, even if we ignore a finite set of points, for each interval
such that
, there is an
which has high density at $J$, even if we require
. Furthermore, roughly speaking, it is possible to choose small
and
s, so that
has arbitrarily high density, and to cover
with these
s. The result follows.
Thank you!
13 January, 2011 at 3:18 am
Terence Tao
Hmm, you’re right; I’ve amended the exercise accordingly.
12 September, 2011 at 5:31 pm
Anonymous
Could someone explain why Em,n is lebesgue measurable (in proof of egorov thm)
23 February, 2016 at 6:45 pm
Joe Li
Since fn is, then so is f (pointwise limit), then so is fn – f, then so si |fn-f|, then so is Em,n.
30 September, 2011 at 11:36 am
Gandhi Viswanathan
Dear Professor Tao,
In Definition 5 above there is a minor typo:
“…to be absolutely integrable of Simp….” should be “…to be absolutely integrable if Simp….”
This typo is also in Definition 1.3.6 of the PDF version of you book on measure theory (which, along with your other books, I am greatly enjoying reading).
[Corrected, thanks – T.]
6 March, 2012 at 5:44 pm
Rex
Stein/Shakarchi’s Real Analysis requires that their simple functions have finite measure support, whereas here you do not adopt any such condition. Is there any advantage gained by making this additional hypothesis for simple functions?
6 March, 2012 at 7:06 pm
Terence Tao
See Remark 1. It does not matter too much what route one takes to the Lebesgue integral, as one has to deal with a fixed set of technical issues, such as horizontal and vertical truncation, additivity, convergence theorems, etc., at some point in the construction. However, the order in which these issues are encountered can depend on what route one selects.
11 December, 2012 at 4:59 pm
Jack
For the Riemann integration, I saw many times that people use
and
interchangeably where
. Is this always justifiable? Also, I saw people use
when they mean
. But
is very ambiguous. It can have at least 3 meanings: 1.
; 2.
; 3.
which is called the principal value. I'm confused which one is supposed to be used. As I understand, 3 is used since 1 and 2 is not always defined and when 1 and 2 are defined, 3 should gives the same result. How can one know whether
means
or the improper integral?
11 December, 2012 at 5:20 pm
Terence Tao
Different texts have different conventions, but personally I use
for the Riemann integral (with the sign convention
) and
for the Lebesgue integral (with the two being of course compatible when integrating continuous functions; see Exercise 17 above).
For improper Riemann integrals such as
, there is no ambiguity if the integrand is either non-negative or absolutely convergent (as then one can use the Lebesgue theory to show that all the different interpretations you mention agree), but otherwise if one wants to be fully rigorous, one does indeed have to specify the precise interpretation of the integral. (Assuming that the author is sufficiently careful, though, this should indeed be the case when encountering improper integrals in a rigorous mathematical context. In looser contexts, such as in calculations done in engineering or physics, these issues might be glossed over, although in most cases one can find some way to make rigorous sense of the integrals appearing in such calculations (though in some cases one has to use the theory of distributions or some regularisation of the integral (e.g. zeta regularisation), rather than the Riemann or Lebesgue theory of integration, to do so).)
11 December, 2012 at 7:15 pm
Anonymous
I’m a little bit confused that in your PCM article “differential forms and integration”, you distinguish
and
in the Riemann integral setting. According to that article,
is a signed definite integral and
is an unsigned definite integral which can be generalized to integration on a measure space. Are these notations in that article interpreted differently from those in your comment?
11 December, 2012 at 8:15 pm
Terence Tao
No, the interpretations in that article and in this blog post are the same (see previous comment about the sign convention). In particular, the signed definite integral and unsigned definite integral are linked in one dimension by the identities
and
for
(when restricting attention to integrating continuous functions, at least).
The quickest way to define integration on forms is indeed by using the connection with the Riemann or Lebesgue integral on a coordinate chart, combined with a suitable partition of unity, as the definition (but then one has to check that the definition is independent of the choice of chart). It is however also possible to define integration of a form without reference to coordinates, for instance by triangulating the manifold and performing a Riemann sum construction directly.
9 January, 2013 at 4:37 pm
Jack
Regarding the proof of Lemma 7, (3)–>(5), why would one need to introduce the
in the proof? Can one use only the Lebesgue measurability of
to do the proof?
9 January, 2013 at 7:22 pm
Terence Tao
No, this is not sufficient, because if one only knows that
for all
, then it could be possible that
is equal to
, rather than being greater than
.
10 January, 2013 at 7:06 am
Jack
In Lemma 7, one has the equivalence between “pointwise convergence” and “almost every convergence”. But generally this is not true. I don’t see why. After all, (3)–>(2) is not obvious (3–>5–>4–>2). What’s the essential reason for this equivalence? To what extend can one generalize it?
10 January, 2013 at 7:14 am
Jack
If one only want (4)–>(2), can one drop the assumption “bounded with finite measure support”? I don’t see where this is used in the proof.
13 January, 2013 at 11:49 am
Jack
Regarding Lemma 7, in the proof of (5.)-(11.) implying (4.), the construction of
seems rather artificial. Could you elaborate the intuition of such construction?
19 January, 2013 at 3:18 pm
Jack
Regarding Exercise 10.8, hint for the unbounded case?
21 January, 2013 at 9:52 am
Jack
It can be done by doing the simple function case and then dealing with “=” by proving “
” and “
” respectively.
I still don’t know how the claim in Exercise 10.3 can fail when
. Is it possible to find a counterexample without using anything related to axiom of choice (Or does one have to use it)?
30 January, 2013 at 9:04 am
Jack
I don’t quite understand the remark after Egorov’s Theorem.
“On a domain such as
, bounded-set locally uniform convergence implies open-neighbourhood locally uniform convergence, but not conversely”
In Remark 6, it is pointed out that these two definition are equivalent and it fails when the domain is where the Heine-Borel theorem does not hold.
In the wikipidia article (http://en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem), the Heine-Borel theorem holds for any subset
. (So it should also hold for
. )
I’m lost… What do I miss here?
30 January, 2013 at 9:51 am
Terence Tao
The notion of a closed set in the relative topology of
is not the same as the notion of a closed set in the larger space
; a subset of
can be closed in the former sense but not in the latter. (For instance,
is closed in
– it contains all its limit points in that space – but not in
). So the Heine-Borel theorem need not be inherited by non-closed subpaces of
; for instance
is closed and bounded in
but fails to be compact.)
30 January, 2013 at 6:04 pm
Jack
OK. It is the relative topology that is used. There is still something I don’t fully understand about the “immediate equivalence” you mentioned in Remark 6.
In
, bounded-set notion implies open-neighborhood definition since one can take a bounded open disk center on any
. But in order to use Heine-Borel to show that the open-neighborhood definition implies the bounded-set one, I think one needs to take closure of a given bounded set first (which renders a compact set, i.e., closed and bounded) and then use the finite open cover properties. It follows that the open-neighborhood definition implies a compact-set definition (replacing “bounded” with “compact”). Hence the bounded-set definition implies the compact-set one. But boundedness would not imply compactness.
I somehow messe up. There must be something wrong here…
(P.S. I thinks the last sentence “since one local uniform convergence in both senses”, which is not in the textbook, in the remark paragraph after Egorov’s theorem, misses some words.)
31 January, 2013 at 6:53 am
Jack
“thinks”—>”think”
Gah, I don’t notice that I asked a silly question until I try to write down everything. Something is true (say, uniform convergence here) on a bounded set is definitely true on a compact set. And thus the compact-set definition is a weaker one than those two (bounded-set and open-neighborhood ones).
4 February, 2013 at 9:22 am
Jack
Regarding Exercise 6, one can first show that it is true for any simple functions and then the bounded unsigned measurable functions (by Exercise 4 where one can use the uniform limit). This is the Trick 3 in the notes (https://terrytao.wordpress.com/2010/10/21/245a-problem-solving-strategies/). But how can one pass back to the general case where
can be unbounded? I don’t see how one can write an unbounded measurable function as a limit/decomposition of a bounded function. (One might use Lemma 7.2 to approximate an unbounded function, but the limit is not uniform…)
19 February, 2013 at 10:20 am
Anonymous
If one uses MCT to prove the linearity of the integration, for example in Corollary 11, should that be viewed as invalid circular proof?
5 May, 2013 at 9:02 pm
Luqing Ye
Dear Prof.Tao,
I think there is a problem to the definition 1(Simple function).I think the domain of a simple function
should not be
,but should be
…
6 May, 2013 at 2:57 am
Shen Zeng
The definition is correct. Where do you see a problem?
6 May, 2013 at 6:29 am
Luqing Ye
I think only when
then the definition would be correct,otherwise,let
,then
would be meaningless.
6 May, 2013 at 6:37 am
Luqing Ye
…Forgive me.I am wrong.The definition is right.……
I have a wrong understanding about indicator function…I didn’t think clearly.
Dear Shen Zeng,last time I have a stupid question about support function,then you appeared.This time I have a stupid question about support function again then you appear again.…
7 May, 2013 at 5:46 am
Luqing Ye
I think a Riemann measurable function can be defined as follows,which is defined in a way similar to Lebesgue measurable function:
1.First,I define Riemann-simple function.This is very similar to Lebesgue-simple function(Definition 1),except that for all
,
is Jordan measurable instead of Lebesgue measurable.And there is another difference,Riemann-simple function is from
to
,but a Lebesgue-simple function is from
to
.
2.Then I define Riemann measurable function to be a function which can be uniformly approximated by Riemann-simple function.
I think a function is Riemann measurable function iff it is Riemann integrable,i.e,it is coherent with the standard definition.
7 May, 2013 at 8:06 am
Luqing Ye
A remark to exercise 4:
It is amazing to see that when
is bounded,then
can be uniformly approximated by a list of simple functions.Although this theorem is just a easy consequence of Hein-Borel finite covering theorem and supremum principle….Regard a simple function as a finite number of points in the vertical direction,then everything will be OK.
This means when
is bounded,we can also define a measurable function
by means of approximation uniformly by a list of simple functions…
So the only purpose of defining a measurable function
by means of approximation pointwise by a list of simple functions is to deal with the case when
is unbounded…
7 May, 2013 at 9:16 am
Luqing Ye
A remark to exercise 6:
I think exercise 6 is a good exercise.Let me guess how Lebesgue created the concept of measurable function.
Firstly,Lebesgue created the Lebesgue measurable set,which is a generalization of Jordan measurable set.I think if there is no such thing as Jordan measurable set in the mathematics,Lebesgue will not discover his measurable set easily.
Then it is evident that we should let a simple function to be measurable.Then,what should we do?
It seems that Lebesgue used an idea which is similar to analytic continuation(I havn’t learned analytic continuation,I only heard this vocabulary,so hope I am not wrong).Because the countable union and countable intersection of Lebesgue measurable set is also Lebesgue measurable,so Lebesgue defined a measurable function to be countable “intersection” and “union” of simple functions,which is exactly a function which can be point wise approximated by simple functions.No less,no more,the concept of measurable function is created exactly suited to our needs!
8 May, 2013 at 12:58 am
Luqing Ye
A remark to Remark 3:
In constructing the L-masurable function
which turns a L-measure 0 set
into L-unmeasurable set
,I think the order preserving property is very important.
To express more explicitly,There is an order preserving bijective map from the Cantor set to
.This order preverving property is essential to the measurability of
.
In fact,this example shows the intuitive procedure that we can project an umeasurable set into a null set,this projection is order preserving and bejective.
10 May, 2013 at 3:08 am
Luqing Ye
A remark to Exercise 10(Basic properties of the lower Lebesgue
integral).
Exercise 10 consists of ten properties.However,I think another
important property should be mentioned,I call this property “Property
11:The principle of correspondence and optimization”,which is invented
by me :-}
Now I describe the principle of correspondence and optimization in
detail.I think the principle of correspondence and optimazation is the
hidden mechanism for property 2,3,5,6,10 in exercise 10.
First I decribe the principle of correspondence and optimization in general.Let
be
be another system.For any state
in system
,there always exists a state
in system
correspoinding to
,then we say that system
is more crude than system
,or,equivalently,system
is more sophisticated than system
.
a system,let
state
If any one of the two systems is more sophisticated than the
other,then we call these two system equi-sophisticated.
Now let me explain the reason why I use the word sophisticate.If for
,there is always a state in
correspoinding to
,there is not neccessarily a state in 
,then we can regard
,suppose that
is equi-sophisticated to
.So the remaining of
has the potential to be more
,just like Pareto improvement in economy.so we call
is more sophisticated than
.
any state in
it,but for any state in
that correspoinding to that state in
system
equi-sophisticated to part of the system
optimized then
To explain my general dicussion ,now we look at a concrete
example.Let’s see property 10 in exercise 10 for example.This property
is stated as follows:
(Reflection) If
is a simple function that is bounded with
.
finite measure support (i.e. it is absolutely integrable), then
Let system
be the set of simple functions
,Let system
be the set of simple functions
. For any simple function
in system
,it is easy to verify that
is a simple function in system
.And for any simple function in system
,it is easy to verify
is a simple function in system
.So system
is
,and system
is more sophisticated than system
,so they are equi-sophisticated.So
.
that
more sophisticated than system
Let’s look at property 5 in exercise 10 for another example.
(Superadditivity)
.
Let system
be the set of simple functions
,let system
be the set of simple functions
.Let
be the set of simple functions
.It is easy to verify that
,
must be in
.But for any
,it
,such that
(Why?),so
.
is not necessarily that there exists
10 May, 2013 at 10:25 am
Luqing Ye
Dear Prof.Tao,
In exercise 10,(9),(Vertical truncation),in the hint,you say:”From Exercise 9 of Notes 1, we have
for any measurable set
. ”
But I can’t see anything special about Exercise 9 of Notes 1(Jordan measurable sets are Lebesgue measurable),Instead,I think it is monotone convergence theorem of measurable sets that plays the central role…
[Corrected, thanks – T.]
11 May, 2013 at 11:11 am
Luqing Ye
A remark to Exercise 10,(8)(Horizontal truncation) .
I think the property of Horizontal truncation should also make use of monetone convergence theorem of measurable sets,which is similar to Vertical truncation.
The only difference is that Horizontal truncation make use of the monetone convergence theorem of measurable sets in the perpendicular to X direction,but the Vertical trucation make use of the monetone convergence theorem of measurable sets in the perpendicular to Y direction.
11 May, 2013 at 10:14 am
Luqing Ye
A generalization to exercise 10,(5),(6).(5) and (6) is stated as
follows:
(Superadditivity)
.
(Subadditivity of upper integral)
.
In order to generalize,let me introduce some new definitions first.Let
be an unsigned function
to be the quantity
(not necessarily measurable). We define the lower unsigned integral
where
ranges over all unsigned measurable functions
that are pointwise bounded by
.
One can also define the upper unsigned Lebesgue integral
With this definition,now I think we can say that
(additivity)
.
(additivity of upper integral)
.
In fact,I think the proof is easy with the help of Exercise 11,and
Exercise 10,(5),(6).
11 May, 2013 at 11:15 am
Luqing Ye
A remark to exercise 11:
Now I answer the question left in exercise 11:
I think that if
is allowed to be unbounded, or is not supported inside a set of finite measure,then the lower Lebesgue integral and upper Lebesgue integral of
still agree.
In order to prove this ,I think I should make use of the Verticle truncation and Horizontal truncation in Exercise 10.
12 May, 2013 at 8:37 am
Luqing Ye
A solution to Exercise 12 (Upper Lebesgue integral and outer Lebesgue measure).
Prove that :
Show that for any set
.
For any set
,let the set
be
.Denote the intersection of all the elements of
to be
.It is easy to verify that
is a measurable set,and
.
And it is easy to verify that
.Done.
13 May, 2013 at 1:13 am
Luqing Ye
A remark to Exercise 14:
I think the property of finite additivity is not neccessary.Finite additivity implies monotonicity:It is monotonicity that is actually used.
For unsigned measurable function
,if for almost every point
,
,then
.
This is the property of monotonicity.I think if monotonnicity replace finite additivity,then the Exercise still holds.
As for the Horizontal and and Vertical truncation,their usage is to extend the L-integral of finite support and bounded function
to infinite in both the Horizontal and Vertical direction.
16 May, 2013 at 8:06 am
Luqing Ye
A question comes to me when I was trying to solve Exercise 16(Linear change of variables).
When
is Riemann integrable,suppose the
on the interval
is
.Let
be a set
,and
.
be an arbitrary real number in the interval
,then
when
,
.
Riemann integration of
of partitioning point of the interval
it is easy to verify that
There is a similar result to Lebesgue integration.When
is Lebesgue measurable,suppose the
on the set
is
.Let
be a sequence of measurable sets,and that
.Now,
,let
.And let
be an arbitrary
.Do we have
when
,
?
Lebesgue integration of
real number on the set
Once I solve this question(Maybe this question is wrong),then Exercise 16 can be easily solved.
16 May, 2013 at 8:11 am
Luqing Ye
I lose an additonal hypothythese,now I add it:
,
.
18 May, 2013 at 1:04 am
Luqing Ye
When I was solving lemma 12(Markov’s inequality),a problem came into my mind,prove or disprove it:
Let
be Lebesgue measurable,and which is pointwise approximated by simple functions
.For any real number
and positive real number
,there exists positive integer
such that
.
22 May, 2013 at 1:21 am
Luqing Ye
Fistly I fix this question.I should have added the additional hypothythese that
should be finite.
Now I disprove this question after it is fixed.The counterexample is in Remark 7—— moving bump.I think moving bump is an interesting example,it is as matter of fact a relationship between the reality and potential.
is approximated by
,this is the reality.But
can be approxiamted by
,this is the potential.
So if I change my question in this way,then it will be true:
Let
be Lebesgue measurable,then there
,such that for any real number
and positive real number
which makes
finite,there exists positive integer k such that
.
exists a sequence of simple functions
which convergent pointwise to
19 May, 2013 at 9:23 am
Luqing Ye
A remark to Lemma 14 (Triangle inequality) .
I enjoy Mr.Tao’s proof of this lemma very much.It is a beautiful,elegant proof!I love this proof very much!
In fact,I prove it in somewhat different way,first,I approximate
by simple functions,then I prove the triangle inequality for simple functions.
But Mr.Tao’s proof is really an amazing proof,in fact,Mr.Tao’s proof has a strong geometric meaning,that is,the smallest path between twp point is a line,because any curve (or polyline) through the two points can project on the line,and the function of projection can only make the length shorter !
Especially I enjoy the formula
This formula strongly reminds me of Fourier transform(Which I only learned from Mr.Tao’s Analysis II),so I think this formula combined with my geometric meaning will give me a very vivid geometric meaning about Fourier transform!
21 May, 2013 at 3:45 am
Luqing Ye
A remark to Theorem 17 (Egorov’s theorem):
I proved this theorem myself.Now I provide my general idea for proving this theorem.For sake of convenience,I take
as a function from
to
.
For any point
,and any
, If
is bounded on
,then let it be.
If
is not bounded on
,then let
be as small
is small enough so that 
,then let it be,in such case,we call
as “not real infinity point”.If
becomes “arbitrarily” small but
still keeps unbounded on
,then just let
be small enough,in this case,we call
as “real infinity point”.
as possible,if when
becomes bounded on
In doing so,we will separate all those open interval in
on which
tends to infinity,and because we let
be small enough,the region we separated has measure as small as
possible(Why?hint:consider the trick give yourself an epsilon of room
https://terrytao.wordpress.com/2009/02/28/tricks-wiki-give-yourself-an-epsilon-of-room/).
Now we consider those open interval on which
is bounded.According to https://terrytao.wordpress.com/2010/09/19/245a-notes-2-the-lebesgue-integral/#comment-228116 It is easy to verify that on those interval,
converges uniformly to
.
So the Egorov’s theorem is proved!
22 May, 2013 at 7:31 am
Luqing Ye
A remark to Egolov’s theorem:
My above comment prove Egolov’s theorem,however,my writing is bad,and
lose several key point.
In my opinion,the key words of proving Egolov’s theorem is “localize
tends to infinity so as to make use of the downward monotone convergence theorem(In order to manage it successfully,we must make use of the regularity of
,because
can be pointwise approximated by a sequence measurable functions,so
itself must be measurable.Measurable function is lovely,it is very regular,i.e,it can be pointwise approximated by simple functions),after that,we use the trick “give yourself an epsilon of room”,to minimize the region on which
tends to infinity.
and minimize”.We localize the region on which
We also localize the region on which
is bounded,to avoid such case as “escape to horizontal infinity” happen.
22 May, 2013 at 5:10 am
Luqing Ye
Dear Professor Tao,
I think Exercise 4 is wrong,it says:
Exercise 4 Let
. Show that
is a bounded unsigned measurable function if and only if
is the uniform limit of bounded simple functions.
A counterexample is the moving bump in Remark 7…
In order to repair,I think
should be bounded in two directions,both horizontal and vertical.
22 May, 2013 at 5:21 am
Luqing Ye
I have to say sorry again because it is my fault.It is the distinction between the potential and reality.
Now I understand that In exercise 4,
is the uniform limit of bounded simple function means there EXISTS a sequence of bounded simple functions which approximated uniformly to
…So the counterexample moving bump is no longer a counterexample,because
can be approximated uniformly by
.
22 May, 2013 at 10:36 am
Luqing Ye
A remark to Lusin’s theorem:
I prove Lusin’s theorem myself,because I don’t like to see proofs provided by other people.
Theorem 18 (Lusin’s theorem) Let
be measurable,and finite almost
. Then there exists a Lebesgue
of measure at most
such that the restriction of
to the complementary set
is continuous on that set.
everywhere, and let
measurable set
Proof:For sake of convenience we ignore the measure zero set,let we
is finite everywhere and measurable.Suppose that
is pointwise approximated by a sequence of simple functions
.According to Egorov’s theorem,for any
,there exists a measurable
such that
,and such that
can be locally uniformly approximated by
.
assume that
given
There is little difference between a simple function and a finite
, where a vanishing function means a function whose
linear combination of indicator functions on boxes.In fact,
support set has measure zero.
So it in fact does no harm to assume that
is a finite linear combination of indicator
functions on at most countable numbers of elementary box(Why it does
no harm?).
Suppose that the support set of
is
,where
,
,
and
are almost disjoint.
is an elementary box who has positive measure.And
Now
,we take
(or
),we cut a very small fraction of
out,which is denoted by
,whose measure is
,after cut
,
is still an elementary box,and such that
.
By using the trick "give yourself an epsilon of room",we can let
as small as possible.
So after this operation,the support set of
becomes
,and we let
be restricted on
.After
is restricted,it can be easily be verified that
is continuous.
It is easy to verify that a if a sequence of continuous function convergent locally uniformly to a function,then that function is also continuous.
So Lusin's theorem is proved!
23 May, 2013 at 5:55 am
Luqing Ye
A remark to Egorov’s theorem:
A generalization of this theorem is stated as follows,but it is not true:
(A false generalization of Egorov’s theorem) Let
be a sequence of measurable functions that converge pointwise almost everywhere to another function
, and let
. Then there exists a Lebesgue measurable set
of measure at most
, such that
converges locally uniformly to
outside of
,and
can not reach
outside of
.
Because if
is from
to
,then in the process of the proof,
will not hold.
9 June, 2014 at 7:32 am
Anonymous
“the upper integral is somewhat problematic when dealing with “improper” integrals of functions that are unbounded or are supported on sets of infinite measure.”
Could you give an example to explain your remark above? (What “problem” are you refering to?)
10 June, 2014 at 7:27 am
Terence Tao
The upper integral of, say,
, will be infinite (if one insists on only using finitely many pieces for the partition). Similarly for
.
12 June, 2014 at 8:33 am
Anonymous
“To obtain the equivalence of (5.) and (6.), observe that
for
and
——————————
instead of
, then it seems that the formula works for
also.
In the first formula, if one uses
In the second formula, why doesn’t it work when
? (the LHS is an empty set and the RHS is an empty union…)
[The formulae work at these endpoints also. -T.]
12 June, 2014 at 1:16 pm
Anonymous
In the first formula, when
, I think it doesn’t work? The left hand side is equal to
but the RHS is an empty intersection (unless one replaces
with
). Or is it a convention that an empty intersection is indeed
?
12 June, 2014 at 2:31 pm
Terence Tao
That would be the appropriate convention in this case: http://math.stackexchange.com/questions/370188/empty-intersection-and-empty-union . In any event, one does not need to consider these endpoint cases for the application at hand.
13 June, 2014 at 8:21 am
Anonymous
I’m trying to get the formula in the proof of Lemma 7:
the set
is equal to
Since
, it suffices to show that
One direction is obvious:
since
for all
. But the other direction is not:
only implies that
Where do I do wrong?
13 June, 2014 at 8:27 am
Terence Tao
The reduction in your “suffices to show that” gives away too much, and is indeed false in general; you should focus on proving the original identity instead. Note that
implies that
.
31 August, 2014 at 12:43 am
Thomas Dybdahl Ahle
Why is the union over
s even necessary? Couldn’t we use just an intersection?
31 August, 2014 at 8:52 am
Terence Tao
The intersection is too small. Consider for instance the example when
and
for all
and
. Then
but
as one can see by considering what happens when
.
31 August, 2014 at 9:39 am
Thomas Dybdahl Ahle
I’m sorry, I didn’t write that clearly. What I meant was that elsewhere in the book (I think) we do limsup’s with just
and not like here where we end up with
.
Equivalently, it seems to me that
By some argument of monotone convergence?
31 August, 2014 at 12:48 pm
Terence Tao
This is not correct; for instance, if
and
, then the right-hand side is
but the left-hand side is empty. However, the right-hand side is equal to
.
31 August, 2014 at 2:59 pm
Thomas Dybdahl Ahle
Ah! The sharp inequality is not preserved by the “limit” so we add an “infinitesimal” on the right hand side using
!
Ingenious. Thank you very much, you are a great inspiration.
22 June, 2014 at 6:29 pm
Anonymous
Do you have a hint for the “only if part” for Exercise 4?
24 June, 2014 at 1:28 pm
Question
For Exercise 4, Using the simple functions (without being truncated vertically, i.e.,
being the largest
less than or equal to
) defined in the last step of the proof of Lemma 7, one has the desired sequence of simple functions.
I don’t see what strategy one should use in
https://terrytao.wordpress.com/2010/10/21/245a-problem-solving-strategies/
It seems that one can take the “finite measure support” out from (4) in Lemma 7. Are simple functions with finite measure support of special interest in application? Isn’t “uniform convergence” preferable?
27 June, 2014 at 12:44 pm
Anonymous
Could you give a hint for Ex 14?
If I don’t understand it wrong, the “uniqueness of the Lebesgue integral” means the following:
Let
be a operator such that
–
when
is simple
for any measurable functions f and g

–
–
–
Show that
for any Lebesgue measurable functions.
29 August, 2014 at 5:08 am
Anonymous
The (real) Lebesgue measurable functions are
measurable by definition. What if one considers the
measurable functions only?
For doing integration of functions
, why is it enough to use Borel measure in the target space?
29 August, 2014 at 8:26 am
Terence Tao
For the purposes of integration theory, the range
and domain
are treated quite differently (in contrast to the more category-theoretic areas of mathematics, in which the domain and range of a morphism are consciously treated on a very equal footing). In particular, the range needs the structure of a complete vector space in order to have any sensible (linear) integration theory. (There are some nonlinear generalisations of the integration concept, but they usually fit better with the more classical Riemann theory of integration than the Lebesgue theory.)
One way to think of a
-measurable function is as the pointwise limit of simple functions (finite linear combinations of indicator functions of measurable sets). This ties in with the underlying intuition of Lebesgue integration as coming from “horizontally” slicing up the graph of a function (as opposed to Riemann integration, which is focused instead on “vertically” slicing up the graph).
Strengthening the measurability requirement by placing the Lebesgue sigma algebra, rather than the Borel sigma algebra, on the range leads to some pathologies, for instance the very natural function
from
to
is now a non-measurable function! (if
is a non-measurable subset of
, then
is a null set and is thus Lebesgue measurable, but not Borel measurable in
.) One can achieve similar pathologies in one-dimension by using an essentially bijective map between, say, the unit interval and the Cantor set.
12 June, 2016 at 5:47 pm
Anonymous
This is a follow up question. When doing probability theory, why Borel measurable functions, namely,
-measurable functions, are used (as random variables), instead of the Lebesgue measurable functions, namely,
.
12 June, 2016 at 9:02 pm
Terence Tao
The property of completeness of a sigma-algebra turns out to not be so useful in probability theory (which tries to ignore the set-theoretic structure of the sample space as much as possible). On the other hand, some technical measure-theoretic constructions, such as disintegration of measure (used to construct conditional expectations) and the Kolmogorov extension theorem are more convenient to state using Borel sigma algebras (particularly if one is working with standard Borel spaces). So there are some slight advantages in probability theory with sticking with Borel sigma algebras throughout, though I doubt there would be much harm in trying to work instead with Lebesgue sigma algebras if one so desired.
15 June, 2016 at 11:51 am
Anonymous
While it is better to use
-measurable functions in probability theory, for what reasons would one need
-measurable functions in general integration theory (for instance, when working in PDE,
-measurable functions are used instead of the
-measurable ones)?
15 June, 2016 at 8:20 pm
Terence Tao
I don’t think it makes much difference in PDE; one often works in function spaces in which one identifies functions which agree almost everywhere, in which case there is no difference between Lebesgue measurable functions and Borel measurable functions. The distinction is certainly important in descriptive set theory, and in certain classical areas of Fourier analysis (e.g. in describing exactly the set of divergence of some Fourier series is). I think there are also some advantages to working in a complete sigma algebra in geometric measure theory, though I can’t think of a concrete such example immediately (other than that it makes it easier to verify that a set is measurable – e.g. it would be enough to show that the outer measure vanishes).
25 September, 2014 at 11:46 am
Integral Bounds In Latex
[…] 245a, notes 2: the lebesgue integral | what’s new […]
28 December, 2015 at 8:44 pm
Anonymous
In Exercise 10 (8), should it be “vertical truncation”? It is not consistent with Notes 3 Exercise 34 (9).
[Notes 3 changed -T.]
23 February, 2016 at 6:55 pm
Joe Li
Dear Prof. Tao and all the readers,
I don’t quite understand some detail of the proof of Lusin’s theorem.
Why we control the measure of A by
instead of
? What is the relation between
and
? And how local uniform limits play any role?
I think that if we control
, then we are almost done. Since
, which is continuous, uniformly converges to
outside of
, then
is continuous outside of
, problem solved. What wrong with this?
24 February, 2016 at 9:58 am
Terence Tao
Sorry, this was a remnant from an earlier version of the post where I also addressed the locally integrable case (but this was pushed to an exercise). The proof has been replaced by the more natural one you indicate.
24 February, 2016 at 12:12 am
k
Hi Terry,
What do you think about this breakthrough in Fourier analysis
http://arxiv.org/abs/1602.05384
18 May, 2016 at 6:45 am
Anonymous
For the linear transformation
, one has from calculus
which looks rather different from Exercise 16. Is this a contradiction?
18 May, 2016 at 7:59 am
Terence Tao
This is the signed definite integral
, which differs from the unsigned definite integral
for
in precisely the fashion indicated. See this previous article of mine for some comparison between the two.
27 September, 2016 at 10:52 am
246A, Notes 2: complex integration | What's new
[…] as the Lebesgue integral. The precise definition of this integral is a little complicated (see e.g. this previous post), but roughly speaking the idea is to approximate by simple functions for some coefficients and […]
23 October, 2016 at 6:54 pm
coupon_clipper
I’m having trouble with Exercise 10.8 and 10.9 (horizontal and vertical truncation properties of the lower Lebesgue integral). In particular, I can’t figure out which other theorems can help me out here since $f$ isn’t even necessarily measurable and therefore neither is
for instance. Can I have a hint?
[
may not be measurable, but the simple functions lying underneath it are. See also the general advice at https://terrytao.wordpress.com/2010/10/21/245a-problem-solving-strategies/ . -T.]
24 October, 2016 at 9:36 am
coupon_clipper
Got it. Thanks so much Terry! Not only do you post this text for free, but you even answer questions on it!
31 August, 2017 at 6:20 am
Anonymous
… horizontal truncation
; alternatively, one can replace
with a bounded variant, such as
.)
Would you elaborate the hint for Exercise 23? I don’t quite understand what that means. Moreover, is
also a bounded variant?
[To obtain Lusin’s theorem for arbitrary measurable
, first apply Lusin’s theorem to a bounded measurable function such as
,
, or
, and figure out how to use the continuous approximation of those functions to obtain a continuous approximation to
. -T]</I