In these notes we quickly review the basics of abstract measure theory and integration theory, which was covered in the previous course but will of course be relied upon in the current course.  This is only a brief summary of the material; of course, one should consult a real analysis text for the full details of the theory.

– Measurable spaces –

Ideally, measure theory on a space X should be able to assign a measure (or “volume”, or “mass”, etc.) to every set in X.  Unfortunately, due to paradoxes such as the Banach-Tarski paradox, many natural notions of measure (e.g. Lebesgue measure) cannot be applied to measure all subsets of X; instead, one must restrict attention to certain measurable subsets of X.  This turns out to suffice for most applications; for instance, just about any “non-pathological” subset of Euclidean space that one actually encounters will be Lebesgue measurable (as a general rule of thumb, any set which does not rely on the axiom of choice in its construction will be measurable).

To formalise this abstractly, we use

Definition 1. A measurable space (X,{\mathcal X}) is a set X, together with a collection {\mathcal X} of subsets of X which form a \sigma-algebra, thus {\mathcal X} contains the empty set and X, and is closed under countable intersections, countable unions, and complements.  A subset of X is said to be measurable with respect to the measurable space if it lies in {\mathcal X}.

A function f: X \to Y from one measurable space (X, {\mathcal X}) to another (Y, {\mathcal Y}) is said to be measurable if f^{-1}(E) \in {\mathcal X} for all E \in {\mathcal Y}.

Remark 1. The class of measurable spaces forms a category, with the measurable functions being the morphisms.   The symbol \sigma stands for “countable union”; cf. \sigma-compact, \sigma-finite, F_\sigma set\diamond

Remark 2. The notion of a measurable space (X,{\mathcal X}) (and of a measurable function) is superficially similar to that of a topological space (X,{\mathcal F}) (and of a continuous function); the topology {\mathcal F} contains \emptyset and X just as the \sigma-algebra {\mathcal X} does, but is now closed under arbitrary unions and finite intersections, rather than countable unions, countable intersections, and complements.  The two categories are linked to each other by the Borel algebra construction, see Example 2 below. \diamond

Example 1. We say that one \sigma-algebra {\mathcal X} on a set X is coarser than another {\mathcal X}' (or that {\mathcal X}' is finer than {\mathcal X}) if {\mathcal X} \subset {\mathcal X'} (or equivalently, if the identity map from (X, {\mathcal X}') to (X, {\mathcal X}) is measurable); thus every set which is measurable in the coarse space is also measurable in the fine space.  The coarsest \sigma-algebra on a set X is the trivial \sigma-algebra \{ \emptyset, X \}, while the finest is the discrete \sigma-algebra 2^X:= \{ E: E \subset X \}. \diamond

Example 2. The intersection \bigwedge_{\alpha \in A} {\mathcal X}_\alpha := \bigcap_{\alpha \in A} {\mathcal X}_\alpha of an arbitrary family ({\mathcal X}_\alpha)_{\alpha \in A} of \sigma-algebras on X is another \sigma-algebra on X.  Because of this, given any collection {\mathcal F} of sets on X we can define the \sigma-algebra {\mathcal B}[{\mathcal F}] generated by {\mathcal F}, defined to be the intersection of all the \sigma-algebras containing {\mathcal F}, or equivalently the coarsest algebra for which all sets in {\mathcal F} are measurable.  (This intersection is non-vacuous, since it will always involve the discrete \sigma-algebra 2^X.)  In particular, the open sets {\mathcal F} of a topological space (X,{\mathcal F}) generate a \sigma-algebra, known as the Borel \sigma-algebra of that space.

We can also define the join \bigvee_{\alpha \in A} {\mathcal X}_\alpha of any family ({\mathcal X}_\alpha)_{\alpha \in A} of \sigma-algebras on X by the formula

\bigvee_{\alpha \in A} {\mathcal X}_\alpha := {\mathcal B}[ \bigcup_{\alpha \in A} {\mathcal X}_\alpha ]. (1)

For instance, the Lebesgue \sigma-algebra {\mathcal L} of Lebesgue measurable sets on a Euclidean space {\Bbb R}^n is the join of the Borel \sigma-algebra {\mathcal B} and of the algebra of null sets and their complements (also called co-null sets). \diamond

Exercise 1. A function f: X \to Y from one topological space to another is said to be Borel measurable if it is measurable once X and Y are equipped with their respective Borel \sigma-algebras.  Show that every continuous function is Borel measurable. (The converse statement, of course, is very far from being true; for instance, the pointwise limit of a sequence of measurable functions, if it exists, is also measurable, whereas the analogous claim for continuous functions is completely false.) \diamond

Remark 3. A function f: {\Bbb R}^n \to {\Bbb C} is said to be Lebesgue measurable if it is measurable from {\Bbb R}^n (with the Lebesgue \sigma-algebra) to {\Bbb C} (with the Borel \sigma-algebra), or equivalently if f^{-1}(B) is Lebesgue measurable for every open ball B in {\Bbb C}.  Note the asymmetry between Lebesgue and Borel here; in particular, the composition of two Lebesgue measurable functions need not be Lebesgue measurable. \diamond

Example 3. Given a function f: X \to Y from a set X to a measurable space (Y, {\mathcal Y}), we can define the pullback f^{-1}({\mathcal Y}) of {\mathcal Y} to be the \sigma-algebra f^{-1}({\mathcal Y}) := \{ f^{-1}(E): E \in {\mathcal Y} \}; this is the coarsest structure on X that makes f measurable.  For instance, the pullback of the Borel \sigma-algebra from [0,1] to {}[0,1]^2 under the map (x,y) \mapsto x consists of all sets of the form E \times [0,1], where E \subset [0,1] is Borel-measurable.

More generally, given a family (f_{\alpha}: X \to Y_\alpha)_{\alpha \in A} of functions into measurable spaces (Y_\alpha, {\mathcal Y}_\alpha), we can form the \sigma-algebra \bigvee_{\alpha \in A} f_\alpha^{-1}({\mathcal Y}_\alpha) generated by the f_\alpha; this is the coarsest structure on X that makes all the f_\alpha simultaneously measurable.\diamond

Remark 4. In probability theory and information theory, the functions f_\alpha: X \to Y_\alpha in Example 3 can be interpreted as observables, and the \sigma-algebra generated by these observables thus captures mathematically the concept of observable information.  For instance, given a time parameter t, one might define the \sigma-algebra {\mathcal F}_{\leq t} generated by all observables for some random process (e.g. Brownian motion) that can be made at time t or earlier; this endows the underlying event space X with an uncountable increasing family of \sigma-algebras. \diamond

Example 4. If E is a subset of a measurable space (Y, {\mathcal Y}), the pullback of {\mathcal Y} under the inclusion map \iota: E \to Y is called the restriction of {\mathcal Y} to E and is denoted {\mathcal Y}\downharpoonright_E.    Thus, for instance, we can restrict the Borel and Lebesgue \sigma-algebras on a Euclidean space {\Bbb R}^n to any subset of such a space. \diamond

Exercise 2. Let M be an n-dimensional manifold, and let (\pi_\alpha: U_\alpha \to V_\alpha) be an atlas of coordinate charts for M, where U_\alpha is an open cover of M and V_\alpha are open subsets of {\Bbb R}^n.  Show that the Borel \sigma-algebra on M is the unique \sigma-algebra whose restriction to each U_\alpha is the pullback via \pi_\alpha of the restriction of the Borel \sigma-algebra of {\Bbb R}^n to V_\alpha. \diamond

Example 5. A function f: X \to A into some index set A will partition X into level sets f^{-1}(\{\alpha\}) for \alpha \in A; conversely, every partition X = \bigcup_{\alpha \in A} E_\alpha of X arises from at least one function f in this manner (one can just take f to be the map from points in X to the partition cell that that point lies in).  Given such an f, we call the \sigma-algebra f^{-1}(2^A) the \sigma-algebra generated by the partition; a set is measurable with respect to this structure if and only if it is the union of some sub-collection \bigcup_{\alpha \in B} E_\alpha of cells of the partition. \diamond

Exercise 3. Show that a \sigma-algebra on a finite set X necessarily arises from a partition X = \bigcup_{\alpha \in A} E_\alpha as in Example 5, and furthermore the partition is unique (up to relabeling).  Thus in the finite world, \sigma-algebras are essentially the same concept as partitions. \diamond

Example 6. Let (X_\alpha, {\mathcal X}_\alpha)_{\alpha \in A} be a family of measurable spaces, then the Cartesian product \prod_{\alpha \in A} X_\alpha has canonical projection maps \pi_\beta: \prod_{\alpha \in A} X_\alpha \to X_\beta for each \beta \in A.  The product \sigma-algebra \prod_{\alpha \in A} {\mathcal X}_\alpha is defined as the \sigma-algebra on \prod_{\alpha \in A} X_\alpha generated by the \pi_\alpha as in Example 3.  \diamond

Exercise 4. Let (X_\alpha)_{\alpha \in A} be an at most countable family of second countable topological spaces.  Show that the Borel \sigma-algebra of the product space (with the product topology) is equal to the product of the Borel \sigma-algebras of the factor spaces. In particular, the Borel \sigma-algebra on {\Bbb R}^n is the product of n copies of the Borel \sigma-algebra on {\Bbb R}.  (The claim can fail when the countability hypotheses are dropped, though in most applications in analysis, these hypotheses are satisfied.)  We caution however that the Lebesgue \sigma-algebra on {\Bbb R}^n is not the product of n copies of the one-dimensional Lebesgue \sigma-algebra, as it contains some additional null sets; however, it is the completion of that product. \diamond

Exercise 5. Let (X,{\mathcal X}) and (Y,{\mathcal Y}) be measurable spaces.  Show that if E is measurable with respect to {\mathcal X} \times {\mathcal Y}, then for every x \in X, the set \{ y \in Y: (x,y) \in E \} is measurable in {\mathcal Y}, and similarly for every y \in Y, the set \{ x \in X: (x,y) \in E \} is measurable in {\mathcal X}.  Thus, sections of Borel-measurable sets are again Borel-measurable.  (The same is not true for Lebesgue-measurable sets.) \diamond

– Measure spaces –

Now we endow measurable spaces with a measure, turning them into measure spaces.

Definition 2 (Measures) A (non-negative) measure \mu on a measurable space (X, {\mathcal X}) is a function \mu: {\mathcal X} \to [0,+\infty] such that \mu(\emptyset)=0, and such that we have the countable additivity property \mu( \bigcup_{n=1}^\infty E_n ) = \sum_{n=1}^\infty \mu(E_n) whenever E_1,E_2,\ldots are disjoint measurable sets.  We refer to the triplet (X,{\mathcal X},\mu) as a measure space.

A measure space (X,{\mathcal X},\mu) is finite if \mu(X) < \infty; it is a probability space if \mu(X)=1 (and then we call \mu a probability measure).  It is \sigma-finite if X can be covered by countably many sets of finite measure.

A measurable set E is a null set if \mu(E)=0.  A property on points x in X  is said to hold for almost every x \in X (or almost surely, for probability spaces) if it holds outside of a null set.  We abbreviate almost every and almost surely as a.e. and a.s. respectively.  The complement of a null set is said to be a co-null set or to have full measure.

Example 7. (Dirac measures)  Given any measurable space (X,{\mathcal X}) and a point x \in X, we can define the Dirac measure (or Dirac mass) \delta_x to be the measure such that \delta_x(E) = 1 when x \in E and \delta_x(E)=0 otherwise. This is a probability measure. \diamond

Example 8. (Counting measure) Given any measurable space (X,{\mathcal X}), we define counting measure \# by defining \#(E) to be the cardinality |E| of E when E is finite, or +\infty otherwise.  This measure is finite when X is finite, and \sigma-finite when X is at most countable.  If X is also finite, we can define normalised counting measure \frac{1}{|E|} \#; this is a probability measure, also known as uniform probability measure on X (especially if we give X the discrete \sigma-algebra). \diamond

Example 9. Any finite non-negative linear combination of measures is again a measure; any finite convex combination of probability measures is again a probability measure. \diamond

Example 10. If f: X \to Y is a measurable map from one measurable space (X,{\mathcal X}) to another (Y, {\mathcal Y}), and \mu is a measure on {\mathcal X}, we can define the push-forward f_* \mu: {\mathcal Y} \to [0,+\infty] by the formula f_* \mu(E) := \mu(f^{-1}(E)); this is a measure on (Y, {\mathcal Y}).  Thus, for instance, f_* \delta_x = \delta_{f(x)} for all x \in X. \diamond

We record some basic properties of measures of sets:

Exercise 6. Let (X,{\mathcal X},\mu) be a measure space.

  1. (Monotonicity) If E \subset F are measurable sets, then \mu(E) \leq \mu(F).  (In particular, any measurable subset of a null set is again a null set.)
  2. (Countable subadditivity) If E_1,E_2,\ldots are a countable sequence of measurable sets, then \mu(\bigcup_{n=1}^\infty E_n) \leq \sum_{n=1}^\infty \mu(E_n).  (Of course, one also has subadditivity for finite sequences.)  In particular, any countable union of null sets is again a null set.
  3. (Monotone convergence for sets) If E_1 \subset E_2 \subset \ldots are measurable, then \mu(\bigcup_{n=1}^\infty E_n) = \lim_{n\to \infty} \mu(E_n).
  4. (Dominated convergence for sets) If E_1 \supset E_2 \supset \ldots are measurable, and \mu(E_1) is finite, then \mu(\bigcap_{n=1}^\infty E_n) = \lim_{n\to \infty} \mu(E_n).  Show that the claim can fail if \mu(E_1) is infinite. \diamond

Exercise 7. A measure space is said to be complete if every subset of a null set is measurable (and is thus again a null set).  Show that every measure space (X, {\mathcal X}, \mu) has a unique minimal complete refinement (X, \overline{{\mathcal X}}, \mu), known as the completion of (X, {\mathcal X}, \mu), and that a set is measurable in \overline{{\mathcal X}} if and only if it is equal almost everywhere to a measurable set in {\mathcal X}.  (The completion of the Borel \sigma-algebra with respect to Lebesgue measure is known as the Lebesgue \sigma-algebra.) \diamond

A powerful way to construct measures on \sigma-algebras {\mathcal X} is to first construct them on a smaller Boolean algebra {\mathcal A} that generates {\mathcal X}, and then extend them via the following result:

Theorem 1. (Carathéodory’s extension theorem, special case)  Let (X, {\mathcal X}) be a measurable space, and let {\mathcal A} be a Boolean algebra (i.e. closed under finite unions, intersections, and complements) that generates {\mathcal X}.  Let \mu: {\mathcal A} \to [0,+\infty] be a function such that

  1. \mu(\emptyset) = 0;
  2. If A_1,A_2,\ldots \in {\mathcal A} are disjoint and \bigcup_{n=1}^\infty A_n \in {\mathcal A}, then \mu( \bigcup_{n=1}^\infty A_n ) = \sum_{n=1}^\infty \mu(A_n).

Then \mu can be extended to a measure \mu: {\mathcal X} \to [0,+\infty] on {\mathcal X}, which we shall also call \mu.

Remark 5. The conditions 1,2 in the above theorem are clearly necessary if \mu has any hope to be extended to a measure on {\mathcal X}.  Thus this theorem gives a necessary and sufficient condition for a function on a Boolean algebra to be extended to a measure.  The extension can easily be shown to be unique when X is \sigma-finite.  \diamond

Proof. (sketch) Define the outer measure \mu_*(E) of any set E \subset X as the infimum of \sum_{n=1}^\infty \mu(A_n), where (A_n)_{n=1}^\infty ranges over all coverings of E by elements in {\mathcal A}.  It is not hard to see that if \mu_* agrees with \mu on {\mathcal A}, so it will suffice to show that it is a measure on {\mathcal X}.

It is easy to check that \mu_* is monotone and countably subadditive (as in parts 1,2 of Exercise 5) on all of 2^X, and assigns zero to \emptyset; thus it is an outer measure in the abstract sense.  But we need to show countable additivity on {\mathcal X}.  The key is to first show the related property

\mu_*(A) = \mu_*(A \cap E) + \mu_*(A \backslash E) (2)

for all A \subset X and E \in {\mathcal X}.  This can first be shown for E \in {\mathcal A}, and then one observes that the class of E that obey (2) for all A is a \sigma-algebra; we leave this as a (moderately lengthy) exercise.

The identity (2) already shows that \mu_* is finitely additive on {\mathcal X}; combining this with countable subadditivity and monotonicity, we conclude that \mu_* is countably additive, as required. \Box

Exercise 8. Let the notation and hypotheses be as in Theorem 1.  Show that given any \varepsilon > 0 and any set E \in {\mathcal X} of finite measure, there exists a set F \in {\mathcal A} which differs from E by a set of measure at most \varepsilon.  If X is \sigma-finite, show that the hypothesis that E have finite measure can be removed.  (Hint: first reduce to the case when X is finite, then show that the class of all E obeying this property is a \sigma-algebra.)  Thus sets in the \sigma-algebra {\mathcal X} “almost” lie in the algebra {\mathcal A}; this is an example of Littlewood’s first principle.   The same statements of course apply for the completion \overline{{\mathcal X}} of {\mathcal X}\diamond

One can use Theorem 1 to construct Lebesgue measure on {\Bbb R} and on {\Bbb R}^n (taking {\mathcal A} to be, say, the algebra generated by half-open intervals or boxes), although the verification of hypothesis 2 of Theorem 2 turns out to be somewhat delicate, even in the one-dimensional case.  But one can at least get the higher-dimensional Lebesgue measure from the one-dimensional one by the product measure construction:

Exercise 9. Let (X_1,{\mathcal X}_1,\mu_1), \ldots, (X_n,{\mathcal X}_n,\mu_n) be a finite collection of measure spaces, and let (\prod_{i=1}^n X_i, \prod_{i=1}^n {\mathcal X}_i) be the product measurable space .  Show that there exists a unique measure \mu on this space such that \mu( \prod_{i=1}^n A_i ) = \prod_{i=1}^n \mu(A_i) for all A_i \in {\mathcal X}_i.  The measure \mu is referred to as the product measure of the \mu_1,\ldots,\mu_n and is denoted \prod_{i=1}^n \mu_i. \diamond

Exercise 10. Let E be a Lebesgue measurable subset of {\Bbb R}^n. and let m be Lebesgue measure.  Establish the inner regularity property

m(E) =\sup \{ \mu(K): K \subset E, \hbox{ compact} \} (3)

and the outer regularity property

{}m(E)=\inf\{\mu(U): E \subset U,\hbox{ op{}en}\} (4).

Combined with the fact that m is locally finite, this implies that m is a Radon measure. \diamond

– Integration –

Now we define integration on a measure space (X, {\mathcal X}, \mu).

Definition 3. (Integration)  Let (X, {\mathcal X}, \mu) be a measure space.

  1. If f: X \to [0,+\infty] is a non-negative simple function (i.e. a measurable function that only takes on finitely many values a_1,\ldots,a_n), we define the integral \int_X f\ d\mu of f to be \int_X f\ d\mu = \sum_{i=1}^n a_i \mu( f^{-1}(\{a_i\}) ) (with the convention that \infty \cdot 0 = 0).  In particular, if f=1_A is the indicator function of a measurable set A, then \int_X 1_A\ d\mu = \mu(A).
  2. If f: X \to [0,+\infty] is a non-negative measurable function, we define the integral \int_X f\ d\mu to be the supremum of \int_X g\ d\mu, where g ranges over all simple functions bounded between 0 and f.
  3. If f: X \to [-\infty,+\infty] is a measurable function, whose positive and negative parts f_+ := \max(f,0), f_- := \max(-f,0) have finite integral, we say that f is absolutely integrable and define \int_X f\ d\mu := \int_X f_+\ d\mu - \int_X f_-\ d\mu.
  4. If f: X \to {\Bbb C} is a measurable function with real and imaginary parts absolutely integrable, we say that f is absolutely integrable and define \int_X f\ d\mu := \int_X \hbox{Re} f\ d\mu + i \int_X \hbox{Im} f\ d\mu.

We will sometimes show the variable of integration, e.g. writing \int_X f(x)\ d\mu(x) for \int_X f\ d\mu, for sake of clarity.

The following results are standard, and the proofs are omitted:

Theorem 2. (Standard facts about integration)  Let (X,{\mathcal X}, \mu) be a measure space.

  1. All the above integration notions are compatible with each other; for instance, if f is both non-negative and absolutely integrable, then definitions 2 and 3 (and 4) agree.
  2. The functional f \mapsto \int_X f\ d\mu is linear over {\Bbb R}^+ for simple functions or non-negative functions, is linear over {\Bbb R} for real-valued absolutely integrable functions, and linear over {\Bbb C} for complex-valued absolutely integrable functions.  In particular, the set of (real or complex) absolutely integrable functions on (X,{\mathcal X},\mu) is a (real or complex) vector space.
  3. A complex-valued measurable function f: X \to {\Bbb C} is absolutely integrable if and only if \int_X |f|\ d\mu < \infty, in which case we have the triangle inequality |\int_X f\ d\mu| \leq \int_X |f|\ d\mu.  Of course, the same claim holds for real-valued measurable functions.
  4. If f: X \to [0,+\infty] is non-negative, then \int_X f\ d\mu \geq 0, with equality holding  if and only if f = 0 a.e.
  5. If one modifies an absolutely integrable function on a set of measure zero, then the new function is also absolutely integrable, and has the same integral as the original function.  Similarly, two non-negative functions that agree a.e. have the same integral.  (Because of this, we can meaningfully integrate functions that are only defined almost everywhere.)
  6. If f: X \to {\Bbb C} is absolutely integrable, then f is finite a.e., and vanishes outside of a \sigma-finite set.
  7. If f: X \to {\Bbb C} is absolutely integrable, and \varepsilon > 0 then there exists a complex-valued simple function g: X \to {\Bbb C} such that \int_X |f-g|\ d\mu \leq \varepsilon.  (This is a manifestation of Littlewood’s second principle.)
  8. (Change of variables formula)  If \phi: X \to Y is a measurable map to another measurable space (Y,{\mathcal Y}), and g: Y \to {\Bbb C}, then we have \int_X g \circ \phi\ d\mu = \int_Y g\ d\phi_* \mu, in the sense that whenever one of the integrals is well defined, then the other is also, and equals the first.

It is also important to note that the Lebesgue integral on {\Bbb R}^n extends the more classical Riemann integral.  As a consequence, many properties of the Riemann integral (e.g. change of variables formula with respect to smooth diffeomorphisms) are inherited by the Lebesgue integral, thanks to various limiting arguments.

We now recall the fundamental convergence theorems relating limits and integration: the first three are for non-negative functions, the last three are for absolutely integrable functions.  They are ultimately derived from their namesakes in Exercise 5 and an approximation argument by simple functions, and the proofs are again omitted.  (They are also closely related to each other, and are in fact largely equivalent.)

Theorem 3. (Convergence theorems) Let (X,{\mathcal X}, \mu) be a measure space.

  1. (Monotone convergence for sequences) If 0 \leq f_1 \leq f_2 \leq \ldots are measurable, then \int_X \lim_{n \to \infty} f_n\ d\mu = \lim_{n \to \infty} \int_X f_n\ d\mu.
  2. (Monotone convergence for series) If f_n: X \to [0,+\infty] are measurable, then \int_X \sum_{n=1}^\infty f_n\ d\mu = \sum_{n=1}^\infty \int_X f_n\ d\mu.
  3. (Fatou’s lemma) If f_n: X \to [0,+\infty] are measurable, then \int_X \liminf_{n \to \infty} f_n\ d\mu \leq \liminf_{n \to \infty} \int_X f_n\ d\mu.
  4. (Dominated convergence for sequences) If f_n: X \to {\Bbb C} are measurable functions converging pointwise a.e. to a limit f, and |f_n| \leq g a.e. for some absolutely integrable g: X \to [0,+\infty], then \int_X \lim_{n \to \infty} f_n\ d\mu = \lim_{n \to \infty} \int_X f_n\ d\mu.
  5. (Dominated convergence for series) If f_n: X \to {\Bbb C} are measurable functions with \sum_n \int_X |f_n|\ d\mu < \infty, then \sum_n f_n(x) is absolutely convergent for a.e. x and \int_X \sum_{n=1}^\infty f_n\ d\mu = \sum_{n=1}^\infty \int_X f_n\ d\mu.
  6. (Egorov’s theorem) If f_n: X \to {\Bbb C} are measurable functions converging pointwise a.e. to a limit f on a subset A of X of finite measure, and \varepsilon > 0, then there exists a set of measure at most \varepsilon, outside of which f_n converges uniformly to f in A.  (This is a manifestation of Littlewood’s third principle.)

Remark 7. As a rule of thumb, if one does not have exact or approximate monotonicity or domination (where “approximate” means “up to an error e whose L^1 norm \int_X |e|\ d\mu goes to zero”), then one should not expect the integral of a limit to equal the limit of the integral in general; there is just too much room for oscillation.  \diamond

Exercise 11. Let f: X \to {\Bbb C} be an absolutely integrable function on a measure space (X,{\mathcal X},\mu).  Show that f is uniformly integrable, in the sense that for every \varepsilon > 0 there exists \delta > 0 such that \int_E |f|\ d\mu \leq \varepsilon whenever E is a measurable set of measure at most \delta.  (The property of uniform integrability becomes more interesting, of course when applied to a family of functions, rather than to a single function.) \diamond

With regard to product measures and integration, the fundamental theorem in this subject is

Theorem 4. (Fubini-Tonelli theorem)  Let (X,{\mathcal X},\mu) and (Y,{\mathcal Y},\nu) be \sigma-finite measure spaces, with product space (X \times Y, {\mathcal X} \times {\mathcal Y}, \mu \times \nu).

  1. (Tonelli theorem) If f: X \times Y \to [0,+\infty] is measurable, then \int_{X\times Y} f\ d\mu \times \nu = \int_X (\int_Y f(x,y)\ d\nu(y))\ d\mu(x) = \int_Y (\int_X f(x,y)\ d\mu(x)) d\nu(y).
  2. (Fubini theorem) If f: X \times Y \to {\Bbb C} is absolutely integrable, then we also have \int_{X\times Y} f\ d\mu \times \nu = \int_X (\int_Y f(x,y)\ d\nu(y))\ d\mu(x) = \int_Y (\int_X f(x,y)\ d\mu(x)) d\nu(y), with the inner integrals being absolutely integrable a.e. and the outer integrals all being absolutely integrable.
  3. If (X,{\mathcal X},\mu) and (Y,{\mathcal Y},\nu) are complete measure spaces, then the same claims hold with the product \sigma-algebra {\mathcal X} \times {\mathcal Y} replaced by its completion.

Remark 8. The theorem fails for non-\sigma-finite spaces, but virtually every measure space actually encountered in “hard analysis” applications will be \sigma-finite.  (One should be cautious, however, with any space constructed using ultrafilters or the first uncountable ordinal.)  It is also important that f obey some measurability in the product space; there exist non-measurable f for which the iterated integrals exist (and may or may not be equal to each other, depending on the properties of f and even on which axioms of set theory one chooses), but the product integral (of course) does not.  \diamond

[Update, Jan 20: statement of Egorov's theorem corrected.]