Product Measures and the Fubini Tonelli Theorems

By [L1] and [A1], the intersection of two measurable rectangles is again a measurable rectangle, and the complement of a measurable rectangle is a finite union of measurable rectangles.

Let $\mathcal{R}$ be the family of finite disjoint unions of measurable rectangles. It contains $\mathrm{\varnothing }$ and $X\times Y$. If $E,F\in \mathcal{R}$, then [A2] disjointifies the finite union $E\cup F$ into finitely many pairwise disjoint measurable rectangles, so $E\cup F\in \mathcal{R}$. Likewise $E\setminus F=E\cap F^c$ belongs to $\mathcal{R}$ by step 1.1. Thus $\mathcal{R}$ is an algebra.

Every measurable rectangle belongs to $\mathcal{R}$, so [L2] gives $\mathcal{A}\otimes \mathcal{B}=\sigma (\text{measurable rectangles})\subseteq \sigma (\mathcal{R})\mathrm{.}$ The reverse inclusion holds because every member of $\mathcal{R}$ is a finite union of measurable rectangles and hence lies in $\mathcal{A}\otimes \mathcal{B}$. Therefore $\sigma (\mathcal{R})=\mathcal{A}\otimes \mathcal{B}$, and $\mathcal{R}$ is an algebra generating the product sigma-algebra.

Fix $x\in X$ and let ${\mathcal{M}}_x$ be the family of sets $F\subseteq X\times Y$ such that $F_x\in \mathcal{B}$. By [A1], every measurable rectangle belongs to ${\mathcal{M}}_x$. By [A2], ${\mathcal{M}}_x$ is a sigma-algebra on $X\times Y$. Since [L1] says $\mathcal{A}\otimes \mathcal{B}$ is generated by the measurable rectangles, [L2] gives $\mathcal{A}\otimes \mathcal{B}\subseteq {\mathcal{M}}_x$. Therefore $E_x\in \mathcal{B}$.

Fix $y\in Y$ and let ${\mathcal{N}}_y$ be the family of sets $F\subseteq X\times Y$ such that $F^y\in \mathcal{A}$. The same argument shows that ${\mathcal{N}}_y$ is a sigma-algebra containing all measurable rectangles, hence all of $\mathcal{A}\otimes \mathcal{B}$. Therefore $E^y\in \mathcal{A}$. Since $x$ and $y$ were arbitrary, every horizontal and vertical section of $E$ is measurable.

Fix $x\in X$ and let $\mathcal{G}$ be a generating family for $\mathcal{B}(\bar{\mathbb{R}})$. For each $U\in \mathcal{G}$, the set $f^{-1}(U)$ is product-measurable, so [L2] and [A1] give $f_x^{-1}(U)=(f^{-1}(U){)}_x\in \mathcal{B}$. By [L1], $f_x$ is therefore $\mathcal{B}$-measurable.

Fix $y\in Y$. The same argument gives $(f^y{)}^{-1}(U)=(f^{-1}(U){)}^y\in \mathcal{A}$ for every generator $U\in \mathcal{G}$, so [L1] implies that $f^y$ is $\mathcal{A}$-measurable. Thus every horizontal and vertical section of $f$ is measurable.

For each $x\in X$, the section $E_x=\{y:y<x\}$ is countable by the choice of $X$, hence measurable for $\mathcal{M}$. For each $y\in X$, the section $E^y=\{x:y<x\}$ has countable complement $\{x:x\le y\}$, hence is cocountable and measurable.

Suppose for contradiction that $E$ were product-measurable for $\mathcal{M}\otimes \mathcal{M}$. Since $\nu (X)=1$, the measure $\nu$ is finite and hence sigma-finite, so [L3] would give ${\int }_X\nu (E_x)d\nu ={\int }_X\nu (E^y)d\nu \mathrm{.}$ But step 1.1 makes $\nu (E_x)=0$ for every $x$ and $\nu (E^y)=1$ for every $y$, so the two sides are $0$ and $1$, a contradiction. Therefore $E$ is not product-measurable, even though all of its sections are measurable. This does not contradict [L1], which proves only the forward implication from product-measurability to section measurability.

Fix $m\ge 1$ and let ${\mathcal{C}}_m$ be the family of sets $F\in \mathcal{A}\otimes \mathcal{B}$ for which $x\mapsto \nu (F_x\cap Y_m)$ is $\mathcal{A}$-measurable. If $F=A\times B$ is a measurable rectangle, then $\nu (F_x\cap Y_m)=\nu (B\cap Y_m){\mathbf{1}}_A(x),$ so $F\in {\mathcal{C}}_m$. By [A2], ${\mathcal{C}}_m$ is a monotone class. Hence [L1] and [L2] imply that every product-measurable set lies in ${\mathcal{C}}_m$.

Applying step 1.1 to the given set $E$ shows that $g_m(x):=\nu (E_x\cap Y_m)$ is measurable for every $m$. Because $Y_m\uparrow Y$, one has $g_m(x)\uparrow \nu (E_x)$ for each $x$, so [L3] gives measurability of $x\mapsto \nu (E_x)$.

The same argument with the finite-measure exhaustion $X_n\uparrow X$ shows that $y\mapsto \mu (E^y)$ is measurable. Therefore both section-measure functions are measurable.

Fix $n,m\ge 1$. Let ${\mathcal{D}}_{n,m}$ be the family of measurable subsets $F\subseteq X_n\times Y_m$ such that ${\int }_{X_n}\nu (F_x)d\mu ={\int }_{Y_m}\mu (F^y)d\nu \mathrm{.}$ If $F=(A\times B)\cap (X_n\times Y_m)$, then $F_x=(B\cap Y_m)$ for $x\in A\cap X_n$ and $\mathrm{\varnothing }$ otherwise, so both integrals equal $\mu (A\cap X_n)\nu (B\cap Y_m)$. Finite additivity gives the same equality for the algebra of [L2].

If $F_k\uparrow F$ inside $X_n\times Y_m$, then [L1] and [L4] give ${\int }_{X_n}\nu ((F_k{)}_x)d\mu \uparrow {\int }_{X_n}\nu (F_x)d\mu ,$ and similarly on $Y_m$. Thus ${\mathcal{D}}_{n,m}$ is a monotone class. By [L2] and [L3], every measurable subset of $X_n\times Y_m$ belongs to ${\mathcal{D}}_{n,m}$.

Put $E_{n,m}:=E\cap (X_n\times Y_m)$. Step 1.2 gives ${\int }_{X_n}\nu ((E_{n,m}{)}_x)d\mu ={\int }_{Y_m}\mu ((E_{n,m}{)}^y)d\nu \mathrm{.}$ Now $(E_{n,m}{)}_x=E_x\cap Y_m$ for $x\in X_n$ and $\mathrm{\varnothing }$ otherwise, so as $n,m\to \mathrm{\infty }$ the two integrands increase pointwise to $\nu (E_x)$ and $\mu (E^y)$.

Applying [L4] on both sides of step 2.1 and then letting $n,m\to \mathrm{\infty }$ gives ${\int }_X\nu (E_x)d\mu ={\int }_Y\mu (E^y)d\nu \mathrm{.}$ This is the claimed equality.

If $E=A\times B$ is a measurable rectangle, then $(A\times B{)}_x=\{\begin{array}{ll}B,& x\in A,\\ \mathrm{\varnothing },& x\mathrm{\notin }A,\end{array}$ so $(\mu \times \nu )(A\times B)={\int }_X\nu ((A\times B{)}_x)d\mu ={\int }_X\nu (B){\mathbf{1}}_A(x)d\mu =\mu (A)\nu (B)\mathrm{.}$ This is the rectangle formula.

Let $E_1,E_2,\dots$ be pairwise disjoint measurable subsets of $X\times Y$, and put $F_N:={\bigcup }_{k\le N}{E}_k$. Then $(F_N{)}_x={\bigcup }_{k\le N}(E_k{)}_x$ is a disjoint union, so $\nu ((F_N{)}_x)={\sum }_{k\le N}\nu ((E_k{)}_x)$ for every $x$. Therefore $(\mu \times \nu )(F_N)={\int }_X{\sum }_{k\le N}\nu ((E_k{)}_x)d\mu \mathrm{.}$ Since $F_N\uparrow {\bigcup }_{k\ge 1}{E}_k$, [L2] gives $(\mu \times \nu )\left({\bigcup }_{k\ge 1}{E}_k\right)={\sum }_{k\ge 1}(\mu \times \nu )(E_k)\mathrm{.}$ Thus $\mu \times \nu$ is a measure.

By [A2] and step 1.1, each rectangle $X_n\times Y_n$ has finite product measure $(\mu \times \nu )(X_n\times Y_n)=\mu (X_n)\nu (Y_n)<\mathrm{\infty },$ and ${\bigcup }_n(X_n\times Y_n)=X\times Y\mathrm{.}$ Hence $\mu \times \nu$ is sigma-finite.

Let $\rho$ be another measure on $\mathcal{A}\otimes \mathcal{B}$ with the same rectangle formula. Step 1.1 shows that $\rho$ and $\mu \times \nu$ agree on the generating pi-system of measurable rectangles, and step 2.1 gives the required sigma-finite exhaustion. Therefore [L3] implies $\rho =\mu \times \nu$ on all of $\mathcal{A}\otimes \mathcal{B}$. This proves existence, the rectangle formula, sigma-finiteness, and uniqueness.

The witness in the Given line satisfies exactly the hypothesis of the displayed claim: $\rho$ and $\tau$ are measures on the same product sigma-algebra and agree on every measurable rectangle.

The same witness also satisfies $\rho (D)=1\mathrm{\ne }0=\tau (D)$, so $\rho \mathrm{\ne }\tau$. Hence the displayed universal uniqueness claim is false. This shows why the sigma-finiteness hypothesis in [L1] is essential.

If $s$ is a nonnegative simple function, write $s={\sum }_{j=1}^m{c}_j{\mathbf{1}}_{E_j}$ with $c_j\ge 0$ and measurable sets $E_j$. Applying [L2] to each indicator ${\mathbf{1}}_{E_j}$ and summing yields $\int sd(\mu \times \nu )={\int }_X\left({\int }_Y{s}_{x}d\nu \right)d\mu ={\int }_Y\left({\int }_X{s}^{y}d\mu \right)d\nu \mathrm{.}$ The inner integral functions are measurable because the same is true for each ${\mathbf{1}}_{E_j}$ and simple combinations preserve measurability.

Choose simple functions $s_n\uparrow f$ by [L3]. Then for each $x$ and $y$ one has $(s_n{)}_x\uparrow f_x$ and $(s_n{)}^y\uparrow f^y$, so [L4] gives ${\int }_Y(s_n{)}_{x}d\nu \uparrow {\int }_Y{f}_{x}d\nu ,{\int }_X(s_n{)}^{y}d\mu \uparrow {\int }_X{f}^{y}d\mu \mathrm{.}$ Applying [L4] once more to the equalities of step 1.1 yields the stated measurability and the equality of all three integrals.

The diagonal $D$ is closed in $[0,1{]}^2$, so $f={\mathbf{1}}_D$ is a nonnegative measurable function; by [A1], it lives on a product space outside the sigma-finite scope of [L1].

For fixed $x\in [0,1]$, [A2] gives $\nu (D_x)=1$, so ${\int }_{[0,1]}\nu (D_x)d\mu (x)={\int }_0^{1}1dx=1.$

For fixed $y\in [0,1]$, [A2] gives $\mu (D^y)=0$, so ${\int }_{[0,1]}\mu (D^y)d\nu (y)={\int }_{[0,1]}0d\nu =0.$

Steps 1.2 and 1.3 give unequal iterated integrals for the same nonnegative measurable function on a non-sigma-finite product space. Hence the displayed universal claim is false, and [L1] cannot be extended by simply deleting sigma-finiteness.

By [L2], the function $\mathrm{\mid }f\mathrm{\mid }$ is nonnegative measurable, and [L1] gives ${\int }_X\left({\int }_Y\mathrm{\mid }{f}_x\mathrm{\mid }d\nu \right)d\mu ={\int }_{X\times Y}\mathrm{\mid }f\mathrm{\mid }d(\mu \times \nu )<\mathrm{\infty }\mathrm{.}$ Therefore ${\int }_Y\mathrm{\mid }{f}_x\mathrm{\mid }d\nu <\mathrm{\infty }$ for $\mu$-almost every $x$, so $f_x\in L^1(\nu )$ for $\mu$-almost every $x$. The same argument with the variables reversed gives $f^y\in L^1(\mu )$ for $\nu$-almost every $y$ and shows that the section integrals of $\mathrm{\mid }f\mathrm{\mid }$ are integrable.

For almost every $x$ from step 1.1, [L4] gives $\mid {\int }_Y{f}_{x}d\nu \mid \le {\int }_Y\mathrm{\mid }{f}_x\mathrm{\mid }d\nu \mathrm{.}$ Define the section integral to be $0$ on the exceptional null set where $f_x\mathrm{\notin }{L}^1(\nu )$. The right-hand side is integrable over $X$ by step 1.1, so this extended section-integral function is integrable. The same convention and conclusion hold for $y\mapsto {\int }_X{f}^{y}d\mu$.

Write $f=u-v+i(p-q)$, where $u,v,p,q\ge 0$ are the positive and negative parts of the real and imaginary parts of $f$. Each of $u,v,p,q$ is integrable because $\mathrm{\mid }u\mathrm{\mid },\mathrm{\mid }v\mathrm{\mid },\mathrm{\mid }p\mathrm{\mid },\mathrm{\mid }q\mathrm{\mid }\le \mathrm{\mid }f\mathrm{\mid }$. Applying [L1] to these four nonnegative functions and recombining with [L3] yields the stated equality of the three integrals.

By [L1], choose a product-measurable function $g$ such that $f=g$ almost everywhere for $\overline{\mu \times \nu }$. Let $N:=\{(x,y):f(x,y)\mathrm{\ne }g(x,y)\}$, so $\overline{\mu \times \nu }(N)=0$. By the completion definition, choose a product-measurable null set $Z$ with $N\subseteq Z$.

Apply the nonnegative case of [L2] to ${\mathbf{1}}_Z$. Since [L2] gives $(\mu \times \nu )(Z)=0$, Tonelli yields ${\int }_X\nu (Z_x)d\mu =0,{\int }_Y\mu (Z^y)d\nu =0.$ Hence $\nu (Z_x)=0$ for $\mu$-almost every $x$ and $\mu (Z^y)=0$ for $\nu$-almost every $y$. Because $N_x\subseteq Z_x$ and $N^y\subseteq Z^y$, the equalities $f_x=g_x$ and $f^y=g^y$ fail only on null sections of the completed factor spaces. For such $x$, the section $f_x$ is almost everywhere equal to the measurable section $g_x$, so [L4] makes $f_x$ $\bar{\mathcal{B}}$-measurable; similarly for $f^y$.

In the nonnegative case, apply Tonelli from [L2] to $g$. Since $f=g$ almost everywhere on the complete product space, the completed integral of $f$ equals that of $g$, and the section integrals agree for the almost-everywhere parameters isolated in step 1.2. This proves part 1.

If $f\in L^1(\overline{\mu \times \nu })$, apply [L1] separately to $\mathrm{Re}f$ and $\mathrm{Im}f$. This gives product-measurable real-valued functions $u,v$ such that $u=\mathrm{Re}f$ and $v=\mathrm{Im}f$ almost everywhere. Put $g:=u+iv$. Then $g$ is product-measurable, $g=f$ almost everywhere, and [L3] applied to $\mathrm{\mid }f\mathrm{\mid }$ and $\mathrm{\mid }g\mathrm{\mid }$ shows $g\in L^1(\mu \times \nu )$. The $L^1$ case of [L2] applies to $g$, step 1.2 transfers the almost-everywhere section measurability from $g$ to $f$, and [L3] transfers the equality of integrals from $g$ to $f$. This proves part 2.

For each $n\ge 1$, the rectangle $Z_n:=\{0\}\times [-n,n]$ satisfies $(\lambda \times \lambda )(Z_n)=\lambda (\{0\})\lambda ([-n,n])=0$ by [L3], so $Z:=\{0\}\times \mathbb{R}={\bigcup }_{n\ge 1}{Z}_n$ is $(\lambda \times \lambda )$-null.

Because $E\subseteq Z$, [L2] puts $E$ in the completed product sigma-algebra, so $f={\mathbf{1}}_E$ is measurable for the completed product.

The section at $0$ is $f_0={\mathbf{1}}_N$, which is not measurable because $N$ is not Lebesgue measurable. Thus the displayed claim fails even though $f$ is measurable for the completed product, and [L1] is sharp.