In the prologue for this course, we recalled the classical theory of Jordan measure on Euclidean spaces . This theory proceeded in the following stages:
- First, one defined the notion of a box and its volume .
- Using this, one defined the notion of an elementary set (a finite union of boxes), and defines the elementary measure of such sets.
- From this, one defined the inner and outer Jordan measures of an arbitrary bounded set . If those measures match, we say that is Jordan measurable, and call the Jordan measure of .
As long as one is lucky enough to only have to deal with Jordan measurable sets, the theory of Jordan measure works well enough. However, as noted previously, not all sets are Jordan measurable, even if one restricts attention to bounded sets. In fact, we shall see later in these notes that there even exist bounded open sets, or compact sets, which are not Jordan measurable, so the Jordan theory does not cover many classes of sets of interest. Another class that it fails to cover is countable unions or intersections of sets that are already known to be measurable:
Exercise 1 Show that the countable union or countable intersection of Jordan measurable sets need not be Jordan measurable, even when bounded.
This creates problems with Riemann integrability (which, as we saw in the preceding notes, was closely related to Jordan measure) and pointwise limits:
Exercise 2 Give an example of a sequence of uniformly bounded, Riemann integrable functions for that converge pointwise to a bounded function that is not Riemann integrable. What happens if we replace pointwise convergence with uniform convergence?
These issues can be rectified by using a more powerful notion of measure than Jordan measure, namely Lebesgue measure. To define this measure, we first tinker with the notion of the Jordan outer measure
of a set (we adopt the convention that if is unbounded, thus now takes values in the extended non-negative reals , whose properties we will briefly review below). Observe from the finite additivity and subadditivity of elementary measure that we can also write the Jordan outer measure as
i.e. the Jordan outer measure is the infimal cost required to cover by a finite union of boxes. (The natural number is allowed to vary freely in the above infimum.) We now modify this by replacing the finite union of boxes by a countable union of boxes, leading to the Lebesgue outer measure of :
thus the Lebesgue outer measure is the infimal cost required to cover by a countable union of boxes. Note that the countable sum may be infinite, and so the Lebesgue outer measure could well equal .
(Caution: the Lebesgue outer measure is sometimes denoted ; this is for instance the case in Stein-Shakarchi.)
Clearly, we always have (since we can always pad out a finite union of boxes into an infinite union by adding an infinite number of empty boxes). But can be a lot smaller:
Example 1 Let be a countable set. We know that the Jordan outer measure of can be quite large; for instance, in one dimension, is infinite, and since has as its closure (see Exercise 18 of the prologue). On the other hand, all countable sets have Lebesgue outer measure zero. Indeed, one simply covers by the degenerate boxes of sidelength and volume zero.
Alternatively, if one does not like degenerate boxes, one can cover each by a cube of sidelength (say) for some arbitrary , leading to a total cost of , which converges to for some absolute constant . As can be arbitrarily small, we see that the Lebesgue outer measure must be zero. We will refer to this type of trick as the trick; it will be used many further times in this course.
From this example we see in particular that a set may be unbounded while still having Lebesgue outer measure zero, in contrast to Jordan outer measure.
As we shall see later in this course, Lebesgue outer measure (also known as Lebesgue exterior measure) is a special case of a more general concept known as an outer measure.
In analogy with the Jordan theory, we would also like to define a concept of “Lebesgue inner measure” to complement that of outer measure. Here, there is an asymmetry (which ultimately arises from the fact that elementary measure is subadditive rather than superadditive): one does not gain any increase in power in the Jordan inner measure by replacing finite unions of boxes with countable ones. But one can get a sort of Lebesgue inner measure by taking complements; see Exercise 18. This leads to one possible definition for Lebesgue measurability, namely the Carathéodory criterion for Lebesgue measurability, see Exercise 17. However, this is not the most intuitive formulation of this concept to work with, and we will instead use a different (but logically equivalent) definition of Lebesgue measurability. The starting point is the observation (see Exercise 5 of the prologue) that Jordan measurable sets can be efficiently contained in elementary sets, with an error that has small Jordan outer measure. In a similar vein, we will define Lebesgue measurable sets to be sets that can be efficiently contained in open sets, with an error that has small Lebesgue outer measure:
Definition 1 (Lebesgue measurability) A set is said to be Lebesgue measurable if, for every , there exists an open set containing such that . If is Lebesgue measurable, we refer to as the Lebesgue measure of (note that this quantity may be equal to ). We also write as when we wish to emphasise the dimension .
(The intuition that measurable sets are almost open is also known as Littlewood’s first principle, this principle is a triviality with our current choice of definitions, though less so if one uses other, equivalent, definitions of Lebesgue measurability.)
As we shall see later, Lebesgue measure extends Jordan measure, in the sense that every Jordan measurable set is Lebesgue measurable, and the Lebesgue measure and Jordan measure of a Jordan measurable set are always equal. We will also see a few other equivalent descriptions of the concept of Lebesgue measurability.
In the notes below we will establish the basic properties of Lebesgue measure. Broadly speaking, this concept obeys all the intuitive properties one would ask of measure, so long as one restricts attention to countable operations rather than uncountable ones, and as long as one restricts attention to Lebesgue measurable sets. The latter is not a serious restriction in practice, as almost every set one actually encounters in analysis will be measurable (the main exceptions being some pathological sets that are constructed using the axiom of choice). In the next set of notes we will use Lebesgue measure to set up the Lebesgue integral, which extends the Riemann integral in the same way that Lebesgue measure extends Jordan measure; and the many pleasant properties of Lebesgue measure will be reflected in analogous pleasant properties of the Lebesgue integral (most notably the convergence theorems).
We will treat all dimensions equally here, but for the purposes of drawing pictures, we recommend to the reader that one sets equal to . However, for this topic at least, no additional mathematical difficulties will be encountered in the higher-dimensional case (though of course there are significant visual difficulties once exceeds ).
The material here is based on Sections 1.1-1.3 of the Stein-Shakarchi text, though it is arranged somewhat differently.
— 1. Two preliminaries: the extended non-negative real axis, and the axiom of choice —
Before we start the main subject of discussion, let us review two basic mathematical tools that we will be using throughout the course: the extended non-negative real axis and the axiom of choice.
The extended non-negative real axis is the non-negative real axis with an additional element adjointed to it, which we label ; we will need to work with this system because many sets (e.g. ) will have infinite measure. Of course, is not a real number, but we think of it as an extended real number. We extend the addition, multiplication, and order structures on to by declaring
for all ,
for all non-zero ,
Most of the laws of algebra for addition, multiplication, and order continue to hold in this extended number system; for instance addition and multiplication are commutative and associative, with the latter distributing over the former, and an order relation is preserved under addition or multiplication of both sides of that relation by the same quantity. However, we caution that the laws of cancellation do not apply once some of the variables are allowed to be infinite; for instance, we cannot deduce from or from . This is related to the fact that the forms and are indeterminate (one cannot assign a value to them without breaking a lot of the rules of algebra). A general rule of thumb is that if one wishes to use cancellation (or proxies for cancellation, such as subtraction or division), this is only safe if one can guarantee that all quantities involved are finite (and in the case of multiplicative cancellation, the quantity being cancelled also needs to be non-zero, of course). However, as long as one avoids using cancellation and works exclusively with non-negative quantities, there is little danger in working in the extended real number system.
We note also that once one adopts the convention , then multiplication becomes upward continuous (in the sense that whenever increases to , and increases to , then increases to ) but not downward continuous (e.g. but ). This asymmetry will ultimately cause us to define integration from below rather than from above, which leads to other asymmetries (e.g. the monotone convergence theorem applies for monotone increasing functions, but not necessarily for monotone decreasing ones).
Remark 1 Note that there is a tradeoff here: if one wants to keep as many useful laws of algebra as one can, then one can add in infinity, or have negative numbers, but it is difficult to have both at the same time. Because of this tradeoff, we will see two overlapping types of measure and integration theory: the non-negative theory, which involves quantities taking values in , and the absolutely integrable theory, which involves quantities taking values in or . For instance, the fundamental convergence theorem for the former theory is the monotone convergence theorem, while the fundamental convergence theorem for the latter is the dominated convergence theorem. Both branches of the theory are important, and both will be covered in later notes.
One important feature of the extended nonnegative real axis is that all sums are convergent: given any sequence , we can always form the sum
as the limit of the partial sums , which may be either finite or infinite. An equivalent definition of this infinite sum is as the supremum of all finite subsums:
Note that when dealing with signed sums, the above rearrangement identity can fail when the series is not absolutely convergent (cf. the Riemann rearrangement theorem).
Exercise 3 If is a collection of numbers such that , show that for all but at most countably many , even if itself is uncountable.
We will rely frequently on the following basic fact (a special case of the Fubini-Tonelli theorem, which we will encounter later in this course):
Informally, Tonelli’s theorem asserts that we may rearrange infinite series with impunity as long as all summands are non-negative.
Proof: We shall just show the equality of the first and second expressions; the equality of the first and third is proven similarly.
We first show that
Let be any finite subset of . Then for some finite , and thus (by the non-negativity of the )
The right-hand side can be rearranged as
which is clearly at most (again by non-negativity of ). This gives
for any finite subset of , and the claim then follows from (1).
It remains to show the reverse inequality
It suffices to show that
for each finite .
Fix . As each is the limit of , the left-hand side is the limit of as . Thus it suffices to show that
for each finite . But the left-hand side is , and the claim follows.
Remark 2 Note how important it was that the were non-negative in the above argument. In the signed case, one needs an additional assumption of absolute summability of on before one is permitted to interchange sums; this is Fubini’s theorem for series, which we will encounter later in this course. Without absolute summability or non-negativity hypotheses, the theorem can fail (consider for instance the case when equals when , when , and otherwise).
Next, we recall the axiom of choice, which we shall be assuming throughout the course:
Axiom 3 (Axiom of choice) Let be a family of non-empty sets , indexed by an index set . Then we can find a family of elements of , indexed by the same set .
This axiom is trivial when is a singleton set, and from mathematical induction one can also prove it without difficulty when is finite. However, when is infinite, one cannot deduce this axiom from the other axioms of set theory, but must explicitly add it to the list of axioms. We isolate the countable case as a particularly useful corollary (though one which is strictly weaker than the full axiom of choice):
Remark 3 The question of how much of real analysis still survives when one is not permitted to use the axiom of choice is a delicate one, involving a fair amount of logic and descriptive set theory to answer. We will not discuss these matters in this course. We will however note a theorem of Gödel that states that any statement that can be phrased in the first-order language of Peano arithmetic, and which is proven with the axiom of choice, can also be proven without the axiom of choice. So, roughly speaking, Gödel’s theorem tells us that for any “finitary” application of real analysis (which includes most of the “practical” applications of the subject), it is safe to use the axiom of choice; it is only when asking questions about “infinitary” objects that are beyond the scope of Peano arithmetic that one can encounter statements that are provable using the axiom of choice, but are not provable without it.
— 2. Properties of Lebesgue outer measure —
We begin by studying the Lebesgue outer measure , which was defined earlier, and takes values in the extended non-negative real axis . We first record three easy properties of Lebesgue outer measure, which we will use repeatedly in the sequel without further comment:
- (Empty set) .
- (Monotonicity) If , then .
- (Countable subadditivity) If is a countable sequence of sets, then . (Hint: Use the axiom of countable choice, Tonelli’s theorem for series, and the trick used previously to show that countable sets had outer measure zero.)
Note that countable subadditivity, when combined with the empty set axiom, gives as a corollary the finite subadditivity property
for any . These subadditivity properties will be useful in establishing upper bounds on Lebesgue outer measure. Establishing lower bounds will often be a bit trickier. (More generally, when dealing with a quantity that is defined using an infimum, it is usually easier to obtain upper bounds on that quantity than lower bounds, because the former requires one to bound just one element of the infimum, whereas the latter requires one to bound all elements.)
Remark 4 Later on in this course, when we study abstract measure theory on a general set , we will define the concept of an outer measure on , which is an assigment of element of to arbitrary subsets of a space that obeys the above three axioms of the empty set, monotonicity, and countable subadditivity; thus Lebesgue outer measure is a model example of an abstract outer measure. On the other hand (and somewhat confusingly), Jordan outer measure will not be an abstract outer measure (even after adopting the convention that unbounded sets have Jordan outer measure ): it obeys the empty set and monotonicity axioms, but is only finitely subadditive rather than countably subadditive. (For instance, the rationals have infinite Jordan outer measure, despite being the countable union of points, each of which have a Jordan outer measure of zero.) Thus we already see a major benefit of allowing countable unions of boxes in the definition of Lebesgue outer measure, in contrast to the finite unions of boxes in the definition of Jordan outer measure, in that finite subadditivity is upgraded to countable subadditivity.
Of course, one cannot hope to upgrade countable subadditivity to uncountable subadditivity: is an uncountable union of points, each of which has Lebesgue outer measure zero, but (as we shall shortly see), has infinite Lebesgue outer measure.
It is natural to ask whether Lebesgue outer measure has the finite additivity property, that is to say that whenever are disjoint. The answer to this question is somewhat subtle: as we shall see later, we have finite additivity (and even countable additivity) when all sets involved are Lebesgue measurable, but that finite additivity (and hence also countable additivity) can break down in the non-measurable case. The difficulty here (which, incidentally, also appears in the theory of Jordan outer measure) is that if and are sufficiently “entangled” with each other, it is not always possible to take a countable cover of by boxes and split the total volume of that cover into separate covers of and without some duplication. However, we can at least recover finite additivity if the sets are separated by some positive distance:
is the distance between and . (Here and in the sequel we use the usual Euclidean metric on .) Then .
Proof: From subadditivity one has , so it suffices to prove the other direction . This is trivial if has infinite Lebesgue outer measure, so we may assume that it has finite Lebesgue outer measure (and then the same is true for and , by monotonicity).
We use the standard “give yourself an epsilon of room” trick. Let . By definition of Lebesgue outer measure, we can cover by a countable family of boxes such that
Suppose it was the case that each box intersected at most one of and . Then we could divide this family into two subfamilies and , the first of which covered , and the second of which covered . From definition of Lebesgue outer measure, we have
summing, we obtain
Since was arbitrary, this gives as required.
Of course, it is quite possible for some of the boxes to intersect both and , particularly if the boxes are big, in which case the above argument does not work because that box would be double-counted. However, observe that given any , one can always partition a large box into a finite number of smaller boxes, each of which has diameter at most , with the total volume of these sub-boxes equal to the volume of the original box. Applying this observation to each of the boxes , we see that given any , we may assume without loss of generality that the boxes covering have diameter at most . In particular, we may assume that all such boxes have diameter strictly less than . Once we do this, then it is no longer possible for any box to intersect both and , and then the previous argument now applies.
In general, disjoint sets need not have a positive separation from each other (e.g. and ). But the situation improves when are closed, and at least one of is compact:
We already know that countable sets have Lebesgue outer measure zero. Now we start computing the outer measure of some other sets. We begin with elementary sets:
Since countable sets have zero outer measure, we note that we have managed to give a proof of Cantor’s theorem that is uncountable. Of course, much quicker proofs of this theorem are available. However, this observation shows that the proof this lemma must somehow use some crucial fact about the real line which is not true for countable subfields of , such as the rationals . In the proof we give here, the key fact about the real line we use is the Heine-Borel theorem, which ultimately exploits the important fact that the reals are complete. In the one-dimensional case , it is also possible to exploit the fact that the reals are connected as a substitute for completeness (note that proper subfields of the reals are neither connected nor complete).
Proof: We already know that , so it suffices to show that .
We first establish this in the case when the elementary set is closed. As the elementary set is also bounded, this allows us to use the powerful Heine-Borel theorem, which asserts that every open cover of has a finite subcover (or in other words, is compact).
We again use the epsilon of room strategy. Let be arbitrary, then we can find a countable family of boxes that cover ,
and such that
We would like to use the Heine-Borel theorem, but the boxes need not be open. But this is not a serious problem, as one can spend another epsilon to enlarge the boxes to be open. More precisely, for each box one can find an open box containing such that (say). The still cover , and we have
As the are open, we may apply the Heine-Borel theorem and conclude that
for some finite . Using the finite subadditivity of elementary measure, we conclude that
Since was arbitrary, the claim follows.
Now we consider the case when the elementary is not closed. Then we can write as the finite union of disjoint boxes, which need not be closed. But, similarly to before, we can use the epsilon of room strategy: for every and every , one can find a closed sub-box of such that (say); then contains the finite union of disjoint closed boxes, which is a closed elementary set. By the previous discussion and the finite additivity of elementary measure, we have
Applying by monotonicity of Lebesgue outer measure, we conclude that
for every . Since was arbitrary, the claim follows.
The above lemma allows us to compute the Lebesgue outer measure of a finite union of boxes. From this and monotonicity we conclude that the Lebesgue outer measure of any set is bounded below by its Jordan inner measure. As it is also bounded above by the Jordan outer measure, we have
for every .
Remark 5 We are now able to explain why not every bounded open set or compact set is Jordan measurable. Consider the countable set , which we enumerate as , let be a small number, and consider the set
This is the union of open sets and is thus open. On the other hand, by countable subadditivity, one has
Finally, as is dense in (i.e. contains ), we have
For small enough (e.g. ), we see that the Lebesgue outer measure and Jordan outer measure of disagree. Using (3), we conclude that the bounded open set is not Jordan measurable. This in turn implies that the complement of in, say, , is also not Jordan measurable, despite being a compact set.
Now we turn to countable unions of boxes. It is convenient to introduce the following notion: two boxes are almost disjoint if their interiors are disjoint, thus for instance and are almost disjoint. As a box has the same elementary measure as its interior, we see that the finite additivity property
holds for almost disjoint boxes , and not just for disjoint boxes. This (and Lemma 6) has the following consequence:
Thus, for instance, itself has an infinite outer measure.
Proof: From countable subadditivity and Lemma 6 we have
so it suffices to show that
But for each natural number , contains the elementary set , so by monotonicity and Lemma 6,
and thus by (4), one has
Letting we obtain the claim.
Remark 6 The above lemma has the following immediate corollary: if can be decomposed in two different ways as the countable union of almost disjoint boxes, then . Although this statement is intuitively obvious and does not explicitly use the concepts of Lebesgue outer measure or Lebesgue measure, it is remarkably difficult to prove this statement rigorously without essentially using one of these two concepts. (Try it!)
Exercise 6 Show that if a set is expressible as the countable union of almost disjoint boxes, then the Lebesgue outer measure of is equal to the Jordan inner measure: , where we extend the definition of Jordan inner measure to unbounded sets in the obvious manner.
Not every set can be expressed as the countable union of almost disjoint boxes (consider for instance the irrationals , which contain no boxes other than the singleton sets). However, there is an important class of sets of this form, namely the open sets:
Proof: We will use the dyadic mesh structure of the Euclidean space , which is a convenient tool for “discretising” certain aspects of real analysis.
Define a closed dyadic cube to be a cube of the form
for some integers . To avoid some technical issues we shall restrict attention here to “small” cubes of sidelength at most , thus we restrict to the non-negative integers, and we will completely ignore “large” cubes of sidelength greater than one. Observe that the closed dyadic cubes of a fixed sidelength are almost disjoint, and cover all of . Also observe that each dyadic cube of sidelength is contained in exactly one “parent” cube of sidelength (which, conversely, has “children” of sidelength ), giving the dyadic cubes a structure analogous to that of a binary tree (or more precisely, an infinite forest of -ary trees). As a consequence of these facts, we also obtain the important dyadic nesting property: given any two closed dyadic cubes (possibly of different sidelength), either they are almost disjoint, or one of them is contained in the other.
If is open, and , then by definition there is an open ball centered at that is contained in , and it is easy to conclude that there is also a closed dyadic cube containing that is contained in . Thus, if we let be the collection of all the dyadic cubes that are contained in , we see that the union of all these cubes is exactly equal to .
As there are only countably many dyadic cubes, is at most countable. But we are not done yet, because these cubes are not almost disjoint (for instance, any cube in will of course overlap with its child cubes). But we can deal with this by exploiting the dyadic nesting property. Let denote those cubes in which are maximal with respect to set inclusion, which means that they are not contained in any other cube in . From the nesting property (and the fact that we have capped the maximum size of our cubes) we see that every cube in is contained in exactly one maximal cube in , and that any two such maximal cubes in are almost disjoint. Thus, we see that is the union of almost disjoint cubes. As is at most countable, the claim follows (adding empty boxes if necessary to pad out the cardinality).
We now have a formula for the Lebesgue outer measure of any open set: it is exactly equal to the Jordan inner measure of that set, or of the total volume of any partitioning of that set into almost disjoint boxes. Finally, we have a formula for the Lebesgue outer measure of an arbitrary set:
Proof: From monotonicity one trivially has
so it suffices to show that
This is trivial for infinite, so we may assume that is finite.
Let . By definition of outer measure, there exists a countable family of boxes covering such that
We use the trick again. We can enlarge each of these boxes to an open box such that . Then the set , being a union of open sets, is itself open, and contains ; and
By countable subadditivity, this implies that
As was arbitrary, we obtain the claim.
Exercise 7 Give an example to show that the reverse statement
is false. (For the corrected version of this statement, see Exercise 15.)
— 3. Lebesgue measurability —
We now define the notion of a Lebesgue measurable set as one which can be efficiently contained in open sets in the sense of Definition 1, and set out their basic properties.
First, we show that there are plenty of Lebesgue measurable sets.
- (i) Every open set is Lebesgue measurable.
- (ii) Every closed set is Lebesgue measurable.
- (iii) Every set of Lebesgue outer measure zero is measurable. (Such sets are called null sets.)
- (iv) The empty set is Lebesgue measurable.
- (v) If is Lebesgue measurable, then so is its complement .
- (vi) If are a sequence of Lebesgue measurable sets, then the union is Lebesgue measurable.
- (vii) If are a sequence of Lebesgue measurable sets, then the intersection is Lebesgue measurable.
Proof: Claim (i) is obvious from definition, as are Claims (iii) and (iv).
To prove Claim (vi), we use the trick. Let be arbitrary. By hypothesis, each is contained in an open set whose difference has Lebesgue outer measure at most . By countable subadditivity, this implies that is contained in , and the difference has Lebesgue outer measure at most . The set , being a union of open sets, is itself open, and the claim follows.
Now we establish Claim (ii). Every closed set is the countable union of closed and bounded sets (by intersecting with, say, the closed balls of radius for ), so by (vi), it suffices to verify the claim when is closed and bounded, hence compact by the Heine-Borel theorem. Note that the boundedness of implies that is finite.
Let . By outer regularity (Lemma 9), we can find an open set containing such that . It suffices to show that .
By monotonicity, the left-hand side is at most , which is in turn at most . Since is finite, we may cancel it and conclude that , as required.
Next, we establish Claim (v). If is Lebesgue measurable, then for every we can find an open set containing such that . Letting be the complement of , we conclude that the complement of contains all of the , and that . If we let , then contains , and from monotonicity , thus is the union of and a set of Lebesgue outer measure zero. But is in turn the union of countably many closed sets . The claim now follows from (ii), (iii), (iv).
Finally, Claim (vii) follows from (v), (vi), and de Morgan’s laws (which work for infinite unions and intersections without any difficulty).
Informally, the above lemma asserts (among other things) that if one starts with such basic subsets of as open or closed sets and then takes at most countably many boolean operations, one will always end up with a Lebesgue measurable set. This is already enough to ensure that the majority of sets that one actually encounters in real analysis will be Lebesgue measurable. (Nevertheless, using the axiom of choice one can construct sets that are not Lebesgue measurable; we will see an example of this later. As a consequence, we cannot generalise the countable closure properties here to uncountable closure properties.)
Remark 7 The properties (iv), (v), (vi) of Lemma 10 assert that the collection of Lebesgue measurable subsets of form a -algebra, which is a strengthening of the more classical concept of a boolean algebra. We will study abstract -algebras in more detail in subsequent notes.
Note how this Lemma 10 is significantly stronger than the counterpart for Jordan measurability (Exercise 6.1 of the prologue), in particular by allowing countably many boolean operations instead of just finitely many. This is one of the main reasons why we use Lebesgue measure instead of Jordan measure.
- (i) is Lebesgue measurable.
- (ii) (Outer approximation by open) For every , one can contain in an open set with .
- (iii) (Almost open) For every , one can find an open set such that . (In other words, differs from an open set by a set of outer measure at most .)
- (iv) (Inner approximation by closed) For every , one can find a closed set contained in with .
- (v) (Almost closed) For every , one can find a closed set such that . (In other words, differs from a closed set by a set of outer measure at most .)
- (vi) (Almost measurable) For every , one can find a Lebesgue measurable set such that . (In other words, differs from a measurable set by a set of outer measure at most .)
(Hint: Some of these deductions are either trivial or very easy. To deduce (i) from (vi), use the trick to show that is contained in a Lebesgue measurable set with , and then take countable intersections to show that differs from a Lebesgue measurable set by a null set.)
Exercise 9 Show that every Jordan measurable set is Lebesgue measurable.
Exercise 10 (Middle thirds Cantor set) Let be the unit interval, let be with the interior of the middle third interval removed, let be with the interior of the middle third of each of the two intervals of removed, and so forth. More formally, write
Let be the intersection of all the elementary sets . Show that is compact, uncountable, and a null set.
Now we look at the Lebesgue measure of a Lebesgue measurable set , which is defined to equal its Lebesgue outer measure . If is Jordan measurable, we see from (3) that the Lebesgue measure and the Jordan measure of coincide, thus Lebesgue measure extends Jordan measure. This justifies the use of the notation to denote both Lebesgue measure of a Lebesgue measurable set, and Jordan measure of a Jordan measurable set (as well as elementary measure of an elementary set).
Lebesgue measure obeys significantly better properties than Lebesgue outer measure, when restricted to Lebesgue measurable sets:
- (Empty set) .
- (Countable additivity) If is a countable sequence of disjoint Lebesgue measurable sets, then .
Using monotonicity, we conclude that
(We can use instead of throughout this argument, thanks to Lemma 10). Sending we obtain
On the other hand, from countable subadditivity one has
and the claim follows.
Next, we handle the case when the are bounded but not necessarily compact. We use the trick. Let . Applying Exercise 8, we know that each is the union of a compact set and a set of outer measure at most . Thus
Finally, from the compact case of this lemma we already know that
while from monotonicity
Putting all this together we see that
for every , while from countable subadditivity we have
The claim follows.
Finally, we handle the case when the are not assumed to be bounded or closed. Here, the basic idea is to decompose each as a countable disjoint union of bounded Lebesgue measurable sets. First, decompose as the countable disjoint union of bounded measurable sets ; for instance one could take the annuli . Then each is the countable disjoint union of the bounded measurable sets for , and thus
by the previous arguments. In a similar vein, is the countable disjoint union of the bounded measurable sets for , and thus
and the claim follows.
From Lemma 11 one of course can conclude finite additivity
whenever are Lebesgue measurable sets. We also have another important result:
- (Upward monotone convergence) Let be a countable non-decreasing sequence of Lebesgue measurable sets. Show that . (Hint: Express as the countable union of the lacunae .)
- (Downward monotone convergence) Let be a countable non-increasing sequence of Lebesgue measurable sets. If at least one of the is finite, show that .
- Give a counterexample to show that the hypothesis that at least one of the is finite in the downward monotone convergence theorem cannot be dropped.
Exercise 12 Show that any map from Lebesgue measurable sets to elements of that obeys the above empty set and countable additivity axioms will also obey the monotonicity and countable subadditivity axioms from Exercise 4, when restricted to Lebesgue measurable sets of course.
Exercise 13 We say that a sequence of sets in converges pointwise to another set in if the indicator functions converge pointwise to .
- Show that if the are all Lebesgue measurable, and converge pointwise to , then is Lebesgue measurable also. (Hint: use the identity or to write in terms of countable unions and intersections of the .)
- (Dominated convergence theorem) Suppose that the are all contained in another Lebesgue measurable set of finite measure. Show that converges to . (Hint: use the upward and downward monotone convergence theorems.)
- Give a counterexample to show that the dominated convergence theorem fails if the are not contained in a set of finite measure, even if we assume that the are all uniformly bounded.
In later notes we will generalise the monotone and dominated convergence theorems to measurable functions instead of measurable sets; these fundamental convergence theorems will be the foundation of much of the rest of the course.
Exercise 14 Let . Show that is contained in a Lebesgue measurable set of measure exactly equal to .
- (i) is Lebesgue measurable with finite measure.
- (ii) (Outer approximation by open) For every , one can contain in an open set of finite measure with .
- (iii) (Almost open bounded) differs from a bounded open set by a set of arbitrarily small Lebesgue outer measure. (In other words, for every there exists a bounded open set such that .)
- (iv) (Inner approximation by compact) For every , one can find a compact set contained in with .
- (v) (Almost compact) differs from a compact set by a set of arbitrarily small Lebesgue outer measure.
- (vi) (Almost bounded measurable) differs from a bounded Lebesgue measurable set by a set of arbitrarily small Lebesgue outer measure.
- (vii) (Almost finite measure) differs from a Lebesgue measurable set with finite measure by a set of arbitrarily small Lebesgue outer measure.
- (viii) (Almost elementary) differs from an elementary set by a set of arbitrarily small Lebesgue outer measure.
- (ix) (Almost dyadically elementary) For every , there exists an integer and a finite union of closed dyadic cubes of sidelength such that .
One can interpret the equivalence of (i) and (ix) in the above exercise as asserting that Lebesgue measurable sets are those which look (locally) “pixelated” at sufficiently fine scales. This will be formalised in later notes with the Lebesgue density theorem.
- (i) is Lebesgue measurable.
- (ii) For every elementary set , one has .
- (iii) For every box , one has .
for any elementary set containing .
- Show that this definition is well defined, i.e. that if are two elementary sets containing , that is equal to .
- Show that , and that equality holds if and only if is Lebesgue measurable.
Exercise 19 Let . Show that the following are equivalent:
- is Lebesgue measurable.
- is a set with a null set removed.
- is the union of a set and a null set.
Remark 9 From the above exercises, we see that when describing what it means for a set to be Lebesgue measurable, there is a tradeoff between the type of approximation one is willing to bear, and the type of things one can say about the approximation. If one is only willing to approximate to within a null set, then one can only say that a measurable set is approximated by a or a set, which is a fairly weak amount of structure. If one is willing to add on an epsilon of error (as measured in the Lebesgue outer measure), one can make a measurable set open; dually, if one is willing to take away an epsilon of error, one can make a measurable set closed. Finally, if one is willing to both add and subtract an epsilon of error, then one can make a measurable set (of finite measure) elementary, or even a finite union of dyadic cubes.
Exercise 20 (Translation invariance) If is Lebesgue measurable, show that is Lebesgue measurable for any , and that .
Exercise 21 (Change of variables) If is Lebesgue measurable, and is a linear transformation, show that is Lebesgue measurable, and that . We caution that if is a linear map to a space of strictly smaller dimension than , then need not be Lebesgue measurable; see Exercise 26.
Exercise 22 Let be natural numbers.
- If and , show that , where denotes -dimensional Lebesgue measure, etc.
- Let , be Lebesgue measurable sets. Show that is Lebesgue measurable, with . (Note that we allow or to have infinite measure, and so one may have to divide into cases or take advantage of the monotone convergence theorem for Lebesgue measure, Exercise 11.)
Exercise 23 (Uniqueness of Lebesgue measure) Show that Lebesgue measure is the only map from Lebesgue measurable sets to that obeys the following axioms:
- (Empty set) .
- (Countable additivity) If is a countable sequence of disjoint Lebesgue measurable sets, then .
- (Translation invariance) If is Lebesgue measurable and , then .
- (Normalisation) .
Hint: First show that must match elementary measure on elementary sets, then show that is bounded by outer measure.
Exercise 24 (Lebesgue measure as the completion of elementary measure) The purpose of the following exercise is to indicate how Lebesgue measure can be viewed as a metric completion of elementary measure in some sense. To avoid some technicalities we will not work in all of , but in some fixed elementary set (e.g. ).
- Let be the power set of . We say that two sets are equivalent if is a null set. Show that this is a equivalence relation.
- Let be the set of equivalence classes of with respect to the above equivalence relation. Define a distance between two equivalence classes by defining . Show that this distance is well-defined (in the sense that whenever and ) and gives the structure of a complete metric space.
- Let be the elementary subsets of , and let be the Lebesgue measurable subsets of . Show that is the closure of with respect to the metric defined above. In particular, is a complete metric space that contains as a dense subset; in other words, is a metric completion of .
- Show that Lebesgue measure descends to a continuous function , which by abuse of notation we shall still call . Show that is the unique continuous extension of the analogous elementary measure function to .
For a further discussion of how measures can be viewed as completions of elementary measures, see these notes.
— 4. Non-measurable sets —
In the previous section we have set out a rich theory of Lebesgue measure, which enjoys many nice properties when applied to Lebesgue measurable sets.
Thus far, we have not ruled out the possibility that every single set is Lebesgue measurable. There is good reason for this: a famous theorem of Solovay asserts that, if one is willing to drop the axiom of choice, there exist models of set theory in which all subsets of are measurable. So any demonstration of the existence of non-measurable sets must use the axiom of choice in some essential way.
That said, we can give an informal (and highly non-rigorous) motivation as to why non-measurable sets should exist, using intuition from probability theory rather than from set theory. The starting point is the observation that Lebesgue sets of finite measure (and in particular, bounded Lebesgue sets) have to be “almost elementary”, in the sense of Exercise 16. So all we need to do to build a non-measurable set is to exhibit a bounded set which is not almost elementary. Intuitively, we want to build a set which has oscillatory structure even at arbitrarily fine scales.
We will non-rigorously do this as follows. We will work inside the unit interval . For each , we imagine that we flip a coin to give either heads or tails (with an independent coin flip for each ), and let be the set of all the for which the coin flip came up heads. We suppose for contradiction that is Lebesgue measurable. Intuitively, since each had a chance of being heads, should occupy about “half” of ; applying the law of large numbers in an extremely non-rigorous fashion, we thus expect to equal .
Moreover, given any subinterval of , the same reasoning leads us to expect that should occupy about half of , so that should be . More generally, given any elementary set in , we should have . This makes it very hard for to be approximated by an elementary set; indeed, a little algebra then shows that for any elementary . Thus is not Lebesgue measurable.
Unfortunately, the above argument is terribly non-rigorous for a number of reasons, not the least of which is that it uses an uncountable number of coin flips, and the rigorous probabilistic theory that one would have to use to model such a system of random variables is too weak to be able to assign meaningful probabilities to such events as “ is Lebesgue measurable”. (For some further discussion of this point, see this post.) So we now turn to more rigorous arguments that establish the existence of non-measurable sets. The arguments will be fairly simple, but the sets constructed are somewhat artificial in nature.
Proposition 12 There exists a subset which is not Lebesgue measurable.
Proof: We use the fact that the rationals are an additive subgroup of the reals , and so partition the reals into disjoint cosets . This creates a quotient group . Each coset of is dense in , and so has a non-empty intersection with . Applying the axiom of choice, we may thus find an element for each . We then let be the collection of all these coset representatives. By construction, .
Also, the different translates are disjoint, because contains only one element from each coset of .
We claim that is not Lebesgue measurable. To see this, suppose for contradiction that was Lebesgue measurable. Then the translates would also be Lebesgue measurable. By countable additivity, we thus have
On the other hand, the sum is either zero (if ) or infinite (if ), leading to the desired contradiction.
Exercise 25 (Outer measure is not finitely additive) Show that there exists disjoint bounded subsets of the real line such that . (Hint: Show that the set constructed in the proof of the above proposition has positive outer measure.)
Remark 10 The above discussion shows that, in the presence of the axiom of choice, one cannot hope to extend Lebesgue measure to arbitrary subsets of while retaining both the countable additivity and the translation invariance properties. If one drops the translation invariant requirement, then this question concerns the theory of measurable cardinals, and is not decidable from the standard ZFC axioms. On the other hand, one can construct finitely additive translation invariant extensions of Lebesgue measure to the power set of by use of the Hahn-Banach theorem to extend the integration functional, though we will not do so here.
(Below are some miscellaneous exercises that were added after the main notes were written; I have not moved it up into a better place so as not to disturb the existing exercise numbering.)
Exercise 27 Define a continuously differentiable curve in to be a set of the form where is a closed interval and is a continuously differentiable function.
- If , show that every continuously differentiable curve has Lebesgue measure zero. (Why is the condition necessary?)
- Conclude that if , then the unit cube cannot be covered by countably many continuously differentiable curves.
We remark that if the curve is only assumed to be continuous, rather than continuously differentiable, then these claims fail, thanks to the existence of space-filling curves.
Exercise 28 (Tonelli’s theorem for series over arbitrary sets) Let be sets (possibly infinite or uncountable), and be a doubly infinite sequence of extended non-negative reals indexed by and . Show that
(Hint: although not strictly necessary, you may find it convenient to first establish the fact that if is finite, then is non-zero for at most countably many .)