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 \(c>0\).