# Fubini’s theorem#

Definition 27

Let $$\mu,\nu$$ be measures on sets $$X,Y$$ respectively and define the set of rectangles to be

$\mathcal R = \{A \subset X : A\ \mu\text{-measurable}\}\otimes \{B\subset Y: B\ \nu\text{-measurable}\}.$

The product measure $$\mu\times \nu$$ on $$X\times Y$$ is defined by

$\mu\times \nu(S) = \inf \sum_{i\in\mathbb N} \mu(A_i)\nu(B_i),$

where the infimum is taken over all countable collections of rectangles $$A_i \times B_i \in \mathcal R$$ with

$S\subset \bigcup_{i\in\mathbb N} A_i \times B_i.$

This is a measure, see Example 62.

Lemma 16

Let $$X,Y$$ be sets and $$\mu,\nu$$ measures on $$X,Y$$ respectively. Then $$\mu\times \nu$$ is equivalently defined by the formula

$\mu\times \nu(S) = \inf \sum_{i\in\mathbb N} \mu(A_i)\nu(B_i),$

where the infimum is taken over all disjoint countable collections of rectangles $$A_i\times B_i \in\mathcal R$$ with

$S\subset \bigcup_{i\in\mathbb N} A_i \times B_i.$

Proof. If $$A,C$$ are $$\mu$$-measurable and $$B,D$$ are $$\nu$$-measurable then

$\begin{split}(A\times B)\setminus (C\times D) &= [(A\setminus C)\times B]\cup [(A\cap C)\times (B\setminus D)]\\ &:= A_1\times B_1 \cup A_2\times B_2\end{split}$

is a decomposition into disjoint rectangles. Since $$A,C$$ and $$B,D$$ are $$\mu$$ and $$\nu$$ measurable respectively,

$\begin{split}\mu(A_1)\nu(B_1) + \mu(A_2)\nu(B_2) &= [\mu(A)-\mu(A\cap C)]\nu(B) + \mu(A\cap C)[\nu(B)-\nu(B\cap D)]\\ &= \mu(A)\nu(B) -\mu(A\cap C)\nu(B\cap D)\\ &\leq \mu(A)\nu(B).\end{split}$

Thus the two formulae agree.

Theorem 22 (Fubini’s theorem)

Let $$X,Y$$ be sets and $$\mu,\nu$$ $$\sigma$$-finite measures on $$X,Y$$ respectively.

1. If $$A$$ is $$\mu$$-measurable and $$B$$ is $$\nu$$-measurable then $$A\times B$$ is $$\mu\times \nu$$-measurable and

$\mu\times \nu(A\times B) = \mu(A)\nu(B).$
2. If $$S$$ is $$\mu\times \nu$$-measurable then

$S^y:= \{x\in X: (x,y)\in S\}$

is $$\mu$$-measurable for $$\nu$$-a.e. $$y\in Y$$, $$y\mapsto \mu(S^y)$$ is $$\nu$$-measurable;

$S_x :=\{y\in Y: (x,y)\in S\}$

is $$\nu$$-measurable for $$\mu$$-a.e. $$x\in X$$, $$x\mapsto \nu(S_x)$$ is $$\mu$$-measurable; and

$\mu\times\nu(S) = \int_Y \mu(S^y)\, \mathrm{d}\nu(y) = \int_X \nu(S_x)\, \mathrm{d}\mu(x).$
3. If $$f\colon X\times Y \to \mathbb R^+$$ is $$\mu\times \nu$$-measurable or $$f\colon X\times Y\to\mathbb R$$ is $$\mu\times \nu$$ integrable then

$\int_{X\times Y}f\, \mathrm{d}\mu\times \nu = \int_X\int_Y f\, \mathrm{d}\nu(y)\, \mathrm{d}\mu(x) = \int_Y\int_X f\, \mathrm{d}\mu(x)\, \mathrm{d}\nu(y).$

Proof. Note that the third point follows from the second by the monotone convergence theorem. We prove the theorem for finite $$\mu,\nu$$.

To begin, let

$\mathcal U:= \left\{\bigcup_{i\in\mathbb N} R_i : R_i \in \mathcal R \text{ pairwise disjoint}\right\}.$

Let $$\mathcal P$$ be the set of $$S\subset X\times Y$$ such that $$y\mapsto \mu(S^y)$$ is $$\nu$$-measurable and for any $$S\in\mathcal P$$ define

$\rho(S):= \int_Y \mu(S^y) \, \mathrm{d}\nu.$

Observe that $$\rho(S)$$ is monotonic in $$S$$.

If $$S=A\times B\in\mathcal R$$ then $$y\mapsto \mu(S^y) =\mu(A)\chi_B$$ is $$\nu$$-measurable and $$\rho(S)=\mu(A)\nu(B)$$. If $$U\in \mathcal U$$ with

$U=\bigcup_{i\in\mathbb N}A_i\times B_i$

a disjoint union, then

$y\mapsto \mu(U^y)= \sum_{i\in\mathbb N} \mu(A_i)\chi_{B_i}$

is a countable sum of $$\nu$$-measurable functions and so $$U\in \mathcal P$$. Moreover,

(35)#$\rho(U)=\int_Y \mu(U^y)\, \mathrm{d}\nu(y)=\sum_{i\in\mathbb N}\mu(A_i)\nu(B_i).$

Thus, for any $$S\subset X\times Y$$,

(36)#$\mu\times\nu(S) = \inf\{\rho(U): S\subset U\in\mathcal U\}.$

To prove the second conclusion, let $$A$$ be $$\mu$$-measurable and $$B$$ be $$\nu$$-measurable. By definition, $$\rho(A\times B) = \mu(A)\nu(B)$$ and, since $$\rho$$ is monotonic, $$\rho(A\times B)\leq \rho(U)$$ whenever $$A\times B \subset U\in\mathcal U$$. Thus, by (36),

$\mu\times \nu(A\times B) = \mu(A)\nu(B).$

Let $$E\subset X\times Y$$ and $$E\subset U\in\mathcal U$$. Observe

$U\cap (A\times B) \quad \text{and} \quad U\setminus (A\times B)$

are disjoint members of $$\mathcal U$$. Therefore, by (35) and (36),

$\begin{split}\rho(U) &= \rho(U\cap (A\times B)) + \rho(V\setminus (A\times B))\\ &\geq \mu\times\nu(E\cap (A\times B)) + \mu\times\nu(E\setminus (A\times B)).\end{split}$

When taking the infimum over all such $$U$$, the left hand side converges to $$\mu\times \nu(E)$$, and so $$A\times B$$ is $$\mu\times\nu$$-measurable. This also implies that all elements of $$\mathcal U$$ are measurable.

Let $$S\subset X\times Y$$ and suppose that $$U_1,U_2,\ldots\in \mathcal U$$ are such that $$\rho(U_i)\to \mu\times\nu(S)$$. Since the intersection of any two rectangles is a rectangle,

$V_i:=U_i\cap U_{i-1}\cap\ldots U_1 \in\mathcal U$

for each $$i\in\mathbb N$$. Moreover, $$\rho(V_i)$$ monotonically decreases to $$\mu\times\nu(S)$$. Let $$W=\bigcap_{i\in\mathbb N}V_i \supset S$$. Note that, for each $$y\in Y$$, $$\mu(V_i^y)$$ monotonically decreases to $$\mu(W^y)$$, so that $$W\in \mathcal P$$. Since $$\mu,\nu$$ are finite, the monotone convergence theorem implies that $$\rho(V_i) \to \rho(W)$$ and hence $$\mu\times \nu(S)=\rho(W)$$. Since $$\mu\times \nu(W) \geq \mu\times\nu(S)$$, the monotonicity of $$\rho$$ implies $$\mu\times\nu(W)=\rho(W)$$.

To prove the first conclusion, if $$S$$ is $$\mu\times\nu$$-measurable, then $$\mu\times\nu(W\setminus S) = 0$$. By (36) there exists $$Z\supset W\setminus U$$ with $$\rho(Z)=0$$. That is, $$\mu(W^y)=\mu(S^y)$$ for $$\nu$$-a.e. $$y\in Y$$, and hence the first conclusion of the second point. Moreover, $$\rho(S)=\rho(W)= \mu\times \nu(S)$$, which concludes the proof.

## Exercises#

Example 62

Let $$X,Y$$ be sets and $$\mu,\nu$$ measures on $$X,Y$$ respectively. Prove that $$\mu\times\nu$$ is a measure on $$X\times Y$$.

Example 63

Let $$X,Y$$ be separable metric spaces. Show that

$\mathcal B(X\times Y) = \Sigma(\mathcal B(X) \otimes \mathcal B(Y)).$

Example 64

Let $$X,Y$$ be separable metric spaces and $$\mu,\nu$$ finite Borel measures on $$X,Y$$ respectively. Show that $$\mu\times\nu|_{\mathcal B(X\times Y)}$$ is the unique countably additive set function on $$\mathcal B(X\times Y)$$ satisfying $$\mu(A\times B)= \mu(A)\times \nu(B)$$ for all $$A\times B\in \mathcal B(X)\otimes \mathcal B(Y)$$.

Example 65

Prove Fubini’s theorem, Example 63 and Example 64 for $$\sigma$$-finite measures $$\mu,\nu$$.

Example 66

Show that the third conclusion of Theorem 22 may fail if

1. $$f\colon X\times Y \to \mathbb R$$ is measurable but not integrable; Hint: consider $$\mu,\nu$$ the counting measure on $$\mathbb N$$. Exploit the “identity”

$(1-1)+(1-1)+ \ldots = 1 + (-1+1)+ (-1+1)+\ldots$
2. $$f\colon X \times Y \to \mathbb R^+$$ is measurable but $$\mu$$ is not $$\sigma$$-finite. Hint: consider $$\mu=\mathcal L^1$$, $$\nu$$ the counting measure on $$\mathbb R$$.

Prove the third conclusion of Theorem 22 for $$f\colon X\times Y \to \mathbb R$$ integrable, even if $$\mu,\nu$$ are not $$\sigma$$-finite.

Example 67

Note in the second conclusion of Theorem 22, we must exclude a set of measure zero. Indeed, if $$\mathcal V$$ is a Vitali set, note that $$\mathcal V\times \{0\} \subset \mathbb R^2$$ is $$\mathcal L^2$$-measurable.

Example 68

Let $$X$$ be a separable metric space and $$f\colon X \to [0,\infty)$$ a Borel function. Prove that

$\int_X f\, \mathrm{d}\mu = \int_{0}^\infty \mu(\{x\in X: f(x)\geq t\})\, \mathrm{d}t.$

Hint: consider

$A=\{(x,t): f(x)\geq t\}.$

Example 69

A measure $$\mu$$ on a metric space $$X$$ is uniformly distributed if there exists a $$g\colon (0,\infty)\to (0,\infty)$$ such that $$\nu(B(x,r))=g(r)$$ for all $$x\in X$$ and $$r>0$$. Let $$\mu,\nu$$ be uniformly distributed Borel regular measures on a separable metric space $$X$$ (with functions $$g$$ and $$h$$ respectively). Let $$U\subset X$$ be open.

1. Observe that, for any $$x\in U$$,

$\lim_{r\to 0} \frac{\nu(U \cap B(x,r))}{h(r)}=1$

for every $$x\in U$$.

2. Deduce that

$\mu(U) \leq \liminf_{r\to 0}h(r)^{-1} \int_U \nu(U \cap B(x,r)) \, \mathrm{d}\mu(x).$
3. Deduce that

$\mu(U) \leq \liminf_{r\to 0}h(r)^{-1} \int_U \mu(U \cap B(y,r)) \, \mathrm{d}\nu(y) = \liminf_{r\to 0} \frac{g(r)}{h(r)}\nu(U).$

Deduce that $$\mu=c\nu$$ for some $$c>0$$.