Proposition 2.2.7379F54715B13

Proof. The product $\sigma $-algebra is generated by the measurable rectangles $A\times B$, and the intersection of two rectangles is a rectangle, so lemma 2.2.6 applies to the family of rectangles.

Step 1: finite measures. Suppose that $\mu $ and $\nu $ are finite. The sets $E\subset X\times Y$ whose sections $E_ x=\{ y:(x,y)\in E\} $ are all measurable form a $\sigma $-algebra containing the rectangles, so sections of product-measurable sets are measurable. The product-measurable sets $E$ for which $x\mapsto \nu (E_ x)$ is measurable include the rectangles, since $\nu ((A\times B)_ x)=\mathbf1_ A(x)\nu (B)$. They are closed under complements, as $\nu ((X\times Y\setminus E)_ x)=\nu (Y)-\nu (E_ x)$, and under countable disjoint unions, so by lemma 2.2.6(1) they are all product-measurable sets.

By monotone convergence (proposition 2.2.3), $\lambda (E)=\int _ X\nu (E_ x)\, \mathrm d\mu (x)$ is a measure with $\lambda (A\times B)=\mu (A)\nu (B)$. If $\lambda '$ is another such measure, then $\lambda '(X\times Y)=\mu (X)\nu (Y)=\lambda (X\times Y)$, so $\lambda '=\lambda $ by lemma 2.2.6(2). In particular, $\lambda $ coincides with the measure $E\mapsto \int _ Y\mu (E^ y)\, \mathrm d\nu (y)$, where $E^ y=\{ x:(x,y)\in E\} $, which is obtained by exchanging the roles of $X$ and $Y$. This proves uniqueness and both integral identities for indicator functions. If $f\geq 0$ is product-measurable, its sections are measurable, because $\{ y:f(x,y)>t\} $ is a section of $\{ f>t\} $. Linearity extends the identities, together with the measurability of the inner integrals, to nonnegative simple functions, and monotone convergence, applied to an increasing sequence of such functions converging to $f$, extends them to $f$.

Step 2: $\sigma $-finite measures. Partition $X$ and $Y$ into countably many measurable sets $X_ m$ and $Y_ n$ of finite measure, and consider the finite measures $\mu _ m(A)=\mu (A\cap X_ m)$ on $X$ and $\nu _ n(B)=\nu (B\cap Y_ n)$ on $Y$. Let $\lambda _{mn}$ be the product of $\mu _ m$ and $\nu _ n$ given by Step 1, and put $\mu \times \nu =\sum _{m,n}\lambda _{mn}$. This is a measure on the product $\sigma $-algebra, and $(\mu \times \nu )(A\times B)=\sum _{m,n}\mu (A\cap X_ m)\nu (B\cap Y_ n)=\mu (A)\nu (B)$. If $\lambda '$ is another measure with this property, then $\lambda '_{mn}(E)=\lambda '(E\cap (X_ m\times Y_ n))$ is a finite measure with $\lambda '_{mn}(A\times B)=\mu _ m(A)\nu _ n(B)$, so $\lambda '_{mn}=\lambda _{mn}$ by Step 1, and $\lambda '=\sum _{m,n}\lambda '_{mn}=\mu \times \nu $.

The integral of a nonnegative measurable function with respect to a countable sum of measures is the sum of the integrals. This holds for indicator functions by definition and for simple functions by linearity; in general it follows by monotone convergence, which also allows us to interchange a limit with a series of nonnegative terms. Applying this to $\mu \times \nu =\sum _{m,n}\lambda _{mn}$, to $\nu =\sum _ n\nu _ n$ and to $\mu =\sum _ m\mu _ m$, and using Step 1 for each $\lambda _{mn}$, we obtain for every nonnegative product-measurable $f$

where $x\mapsto \int _ Yf(x,y)\, \mathrm d\nu (y)=\sum _ n\int _ Yf(x,y)\, \mathrm d\nu _ n(y)$ is measurable as a countable sum of measurable functions. The other iterated integral is treated in the same way, with the roles of $X$ and $Y$ exchanged.

Now let $\int |f|\, \mathrm d(\mu \times \nu )<\infty $. By the nonnegative case applied to $|f|$, the measurable function $x\mapsto \int _ Y|f(x,y)|\, \mathrm d\nu (y)$ is integrable, so it is finite outside a measurable $\mu $-null set $N$. For $x\notin N$ the section $f(x,\cdot )$ is integrable, and its integral is the corresponding combination of the integrals of the sections of the positive and negative parts of $\operatorname{Re}f$ and $\operatorname{Im}f$. These four functions have finite integrals with respect to $\mu \times \nu $; defining the inner integral to be zero on $N$ and applying the nonnegative case to each of them, we obtain the first equality for $f$. The second follows by exchanging the roles of $X$ and $Y$. $\square$