Proof of Tonelli and Fubini Theorems

We use the $[0,\infty]$-valued measurability and integral of Lebesgue Integral of a Nonnegative Measurable Function and the $[0,\infty]$ conventions of Measure, Measure Space, and Probability Measure. Fix disjoint decompositions $X=\bigcup_iC_i$, $Y=\bigcup_nD_n$ with $\mu(C_i)<\infty$, $\nu(D_n)<\infty$, and finite measures $\nu_n(B)=\nu(B\cap D_n)$, $\mu_i(A)=\mu(A\cap C_i)$, as in Step 0 of the proof of Existence and Uniqueness of the Product Measure. For $E\in\mathcal{F}\otimes\mathcal{G}$ and $x\in X$, $y\in Y$, write $E_x=\{y:(x,y)\in E\}$ and $E^y=\{x:(x,y)\in E\}$.

Step 1 (indicator Tonelli). By Steps 1 and 2 of the proof of Existence and Uniqueness of the Product Measure: $E_x\in\mathcal{G}$ for every $x$, each $x\mapsto\nu_n(E_x)$ is real-valued $\mathcal{F}$-measurable, and the measure constructed there, which by the uniqueness clause of Existence and Uniqueness of the Product Measure equals $\mu\otimes\nu$, satisfies

$$(\mu\otimes\nu)(E)=\sum_n\int_X\nu_n(E_x)\,d\mu(x).$$

Now $\nu(E_x)=\sum_n\nu_n(E_x)$ by countable additivity along the disjoint $D_n$, so $x\mapsto\nu(E_x)$ is $[0,\infty]$-valued measurable (a nondecreasing pointwise supremum of measurable partial sums, using $\{\sup_j u_j>t\}=\bigcup_j\{u_j>t\}$), and by Monotone Convergence Theorem together with claim 1 of Linearity and Monotonicity of the Lebesgue Integral (finite sums pull out; the partial sums increase),

$$(\mu\otimes\nu)(E)=\int_X\nu(E_x)\,d\mu(x).$$

Symmetrically, the set function $E\mapsto\sum_i\int_Y\mu_i(E^y)\,d\nu(y)$ is, by the same arguments with the roles of the factors exchanged, a measure on $\mathcal{F}\otimes\mathcal{G}$ assigning $\mu(A)\nu(B)$ to every measurable rectangle; the uniqueness clause of Existence and Uniqueness of the Product Measure identifies it with $\mu\otimes\nu$ as well, giving

$$(\mu\otimes\nu)(E)=\int_Y\mu(E^y)\,d\nu(y).$$

Step 2 (sections of functions). Let $f:X\times Y\to[0,\infty]$ be $\mathcal{F}\otimes\mathcal{G}$-measurable. For fixed $x$ and real $t$, $\{y:f_x(y)>t\}=(\{f>t\})_x\in\mathcal{G}$ by Step 1 sections; hence $f_x$ is $\mathcal{G}$-measurable, and symmetrically for $f^y:x\mapsto f(x,y)$. This proves the Sections claim.

Step 3 (Tonelli). For indicators $f=\mathbf{1}_E$ the three quantities in the display of the statement coincide by Step 1 (note $\int_Y(\mathbf{1}_E)_x\,d\nu=\nu(E_x)$). For nonnegative simple $f$ the identity follows by linearity (claim 1 of Linearity and Monotonicity of the Lebesgue Integral) applied on $(Y,\mathcal{G},\nu)$ for each fixed $x$, then on $(X,\mathcal{F},\mu)$, then on the product; sums of measurable functions are measurable (rational-union argument with density of the rationals). For general $f$, let $\varphi_L$ be the dyadic staircase functions of Step 0(b) of the proof of Linearity and Monotonicity of the Lebesgue Integral, extended by $\varphi_L(\infty)=L$; then $\varphi_L\circ f$ are nonnegative simple, nondecreasing in $L$, with $\varphi_L\circ f\uparrow f$ pointwise (at points where $f=\infty$ the values $L$ increase to $\infty$). For each fixed $x$, $(\varphi_L\circ f)_x\uparrow f_x$, so by Monotone Convergence Theorem on $(Y,\mathcal{G},\nu)$,

$$\int_Yf_x\,d\nu=\sup_L\int_Y(\varphi_L\circ f)_x\,d\nu;$$

the right side is a nondecreasing supremum of $\mathcal{F}$-measurable functions of $x$ (simple case), so $x\mapsto\int_Yf_x\,d\nu$ is measurable, and by Monotone Convergence Theorem on $(X,\mathcal{F},\mu)$ (the inner integrals are nondecreasing in $L$ by monotonicity) and on the product space,

$$\int_X\Bigl(\int_Yf_x\,d\nu\Bigr)d\mu=\sup_L\int_X\Bigl(\int_Y(\varphi_L\circ f)_x\,d\nu\Bigr)d\mu=\sup_L\int_{X\times Y}\varphi_L\circ f\,d(\mu\otimes\nu)=\int_{X\times Y}f\,d(\mu\otimes\nu).$$

The other order is symmetric. This proves Tonelli.

Step 4 (Fubini). Let $f:X\times Y\to\mathbb{R}$ be integrable with respect to $\mu\otimes\nu$. Applying Step 3 to $|f|$, the $[0,\infty]$-valued measurable function $G(x)=\int_Y|f_x|\,d\nu$ satisfies $\int_XG\,d\mu=\int|f|\,d(\mu\otimes\nu)<\infty$. Let $N=\{x:G(x)=\infty\}=\bigcap_L\{G>L\}\in\mathcal{F}$. If $\mu(N)>0$ then, since $G\ge L\,\mathbf{1}_N$ pointwise for every real $L$, monotonicity and the simple-function integral would give $\int_XG\,d\mu\ge L\,\mu(N)$ for all $L$, contradicting finiteness; so $\mu(N)=0$.

For $x\notin N$: $(f^{\pm})_x=(f_x)^{\pm}$ are $\mathcal{G}$-measurable (Step 2 applied to $f^{\pm}$) with $\int_Y(f^{\pm})_x\,d\nu\le G(x)<\infty$, so $f_x$ is $\nu$-integrable. Define $u^{\pm}(x)=\mathbf{1}_{X\setminus N}(x)\int_Y(f^{\pm})_x\,d\nu$; these are real-valued (finite off $N$, zero on $N$), measurable (for $t\ge0$, $\{u^{\pm}>t\}=(X\setminus N)\cap\{\int_Y(f^{\pm})_x\,d\nu>t\}$, and for $t<0$ the set is $X$), and the function $H$ of the statement (equal to $\int_Yf_x\,d\nu$ off $N$, zero on $N$) is $H=u^{+}-u^{-}$, with $|H|\le\mathbf{1}_{X\setminus N}\,G\le G$; hence $H$ is $\mu$-integrable. Moreover $u^{\pm}$ and $x\mapsto\int_Y(f^{\pm})_x\,d\nu$ differ only on $N$, and a $[0,\infty]$-valued measurable function $h$ supported on the $\mu$-null set $N$ has $\int h\,d\mu=0$ (its dyadic approximations are bounded by multiples of $\mathbf{1}_N$, whose integrals vanish; take the supremum). Hence, by Linearity and Monotonicity of the Lebesgue Integral and Step 3 applied to $f^{\pm}$,

$$\int_XH\,d\mu=\int_Xu^{+}\,d\mu-\int_Xu^{-}\,d\mu=\int f^{+}\,d(\mu\otimes\nu)-\int f^{-}\,d(\mu\otimes\nu)=\int_{X\times Y}f\,d(\mu\otimes\nu),$$

the last equality being the definition of the integral of an integrable function. The statement in the other order follows symmetrically. $\blacksquare$