# Fubini’s theorem

## Contents

# Fubini’s theorem#

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

The *product measure* \(\mu\times \nu\) on \(X\times Y\) is
defined by

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

This is a measure, see Example 62.

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

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

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

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

Thus the two formulae agree.

(Fubini’s theorem)

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

If \(A\) is \(\mu\)-measurable and \(B\) is \(\nu\)-measurable then \(A\times B\) is \(\mu\times \nu\)-measurable and

\[\mu\times \nu(A\times B) = \mu(A)\nu(B).\]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

\[\mu\times\nu(S) = \int_Y \mu(S^y)\, \mathrm{d}\nu(y) = \int_X \nu(S_x)\, \mathrm{d}\mu(x).\]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

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

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

a disjoint union, then

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

Thus, for any \(S\subset X\times Y\),

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),

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

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

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,

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#

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

Let \(X,Y\) be separable metric spaces. Show that

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 \(A\times B\in \mathcal B(X)\otimes \mathcal B(Y)\).

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

Show that the third conclusion of Theorem 22 may fail if

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

\[(1-1)+(1-1)+ \ldots = 1 + (-1+1)+ (-1+1)+\ldots\]\(f\colon X \times Y \to \mathbb R^+\) 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.

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.

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

Hint: consider

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.

Observe that, for any \(x\in U\),

\[\lim_{r\to 0} \frac{\nu(U \cap B(x,r))}{h(r)}=1\]for every \(x\in U\).

Deduce that

\[\mu(U) \leq \liminf_{r\to 0}h(r)^{-1} \int_U \nu(U \cap B(x,r)) \, \mathrm{d}\mu(x).\]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 \(c>0\).