Fubini’s theorem
Contents
Fubini’s theorem#
Definition 27
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.
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
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.
Theorem 22 (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#
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
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
\(f\colon X\times Y \to \mathbb R\) 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.
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
Hint: consider
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.
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 \(c>0\).