Theorem 15.8.3 (Fubini-Tonelli Theorem). Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be $\sigma$-finite measure spaces, then
For any $f \in L^{+}(X \times Y)$, $[x \mapsto \int f(x, y)\nu(dy)] \in L^{+}(X)$, $[y \mapsto \int f(x, y)\mu(dx)] \in L^{+}(Y)$, and
\begin{align*}\int_{X \times Y}f(z)\mu \otimes \nu(dz)&= \int_{X}\int_{Y} f(x, y)\nu(dy)\mu(dx) \\&= \int_{Y}\int_{X} f(x, y)\mu(dx)\nu(dy)\end{align*}For any normed space $E$ and $f \in L^{1}(X \times Y; E)$,
For almost every $x \in X$, $f(x, \cdot) \in L^{1}(Y; E)$. For almost every $y \in Y$, $f(\cdot, y) \in L^{1}(X; E)$.
$[x \mapsto \int f(x, y)\nu(dy)] \in L^{1}(X; E)$ and $[y \mapsto \int f(x, y)\mu(dx)] \in L^{1}(Y; E)$.
- \begin{align*}\int_{X \times Y}f(z)\mu \otimes \nu(dz)&= \int_{X}\int_{Y} f(x, y)\nu(dy)\mu(dx) \\&= \int_{Y}\int_{X} f(x, y)\mu(dx)\nu(dy)\end{align*}
Proof. (1): First suppose that $\mu$ and $\nu$ are both finite. Let $\alg \subset \cm \otimes \cn$ be the collection of sets whose indicator functions satisfy (1), then $\alg$ contains all rectangles with finite measure. By linearity of the integral, for any $E, F\in \alg$, $E \setminus F \in \alg$. For any $\seq{E_n}\subset \alg$ and $E \in \alg$ such that $E_{n} \upto E$, by the Monotone Convergence Theorem,
so $\alg$ is a $\lambda$-system. By Dynkin’s $\pi$-$\lambda$ Theorem, $\alg = \cm \otimes \cn$.
Now suppose that $\mu$ and $\nu$ are $\sigma$-finite, then $\alg$ contains all sets in $\cm \otimes \cn$ with finite measure. Since $\mu$ and $\nu$ are $\sigma$-finite, there exists rectangles $\seq{A_n \times B_n}\subset \alg$ such that $\mu \otimes \nu(A_{n} \times B_{n})< \infty$ for all $n \in \natp$ and $A_{n} \times B_{n} \upto X \times Y$. Let $A \in \cm \otimes \cn$, then by the Monotone Convergence Theorem,
Therefore $\alg = \cm \otimes \cn$.
Now let $F \subset L^{+}(X \times Y)$ be the set of functions satisfying (1), then $F \supset \Sigma^{+}(X, \cm)$ by linearity. Let $f \in L^{+}(X \times Y)$, then by Proposition 18.5.5, there exists $\seq{\phi_n}\subset \Sigma^{+}(X \times Y)$ such that $\phi_{n} \upto f$. In which case, by the Monotone Convergence Theorem, (1) also holds for $f$.$\square$