Fubini’s Theorem

Informally, for a nonnegative or integrable function $f:{\Omega }_1\times {\Omega }_2\to \mathbb{R}$, we have

$${\int }_{{\Omega }_1\times {\Omega }_2}f\mathrm{d}\mu (x,y)={\int }_{{\Omega }_1}{\int }_{{\Omega }_2}f\mathrm{d}{\mu }_2(y)\mathrm{d}{\mu }_1(x)={\int }_{{\Omega }_2}{\int }_{{\Omega }_1}f\mathrm{d}{\mu }_1(x)\mathrm{d}{\mu }_2(y).$$

Nonnegative Functions

Let $f:{\Omega }_1\times {\Omega }_2\to \mathbb{R}$ be a nonnegative function and $\mu ={\mu }_1\otimes {\mu }_2$ be a $\sigma$-finite¹ product measure on ${\Omega }_1\times {\Omega }_2$. Then

• $f({\omega }_1,{\omega }_2)$ is a measurable function of ${\omega }_1$ and ${\omega }_2$.
• $f_1(x_1)={\int }_{{\Omega }_2}f(x_1,x_2)d{\mu }_2(x_2)$ is measurable on ${\Omega }_1$.
• $f_2(x_2)={\int }_{{\Omega }_1}f(x_1,x_2)d{\mu }_1(x_1)$ is measurable on ${\Omega }_2$.

And we have

$${\int }_{{\Omega }_1\times {\Omega }_2}fd\mu ={\int }_{{\Omega }_1}{f}_{1}d{\mu }_1={\int }_{{\Omega }_2}{f}_{2}d{\mu }_2.$$

Integrable Functions

Let $f:{\Omega }_1\times {\Omega }_2\to \mathbb{R}$ be an integrable function with respect to a $\sigma$-finite measure $\mu ={\mu }_1\otimes {\mu }_2$.² Then

• $f({\omega }_1,{\omega }_2)$ is an integrable function of ${\omega }_1$ for almost all ${\omega }_2$.
• $f({\omega }_1,{\omega }_2)$ is an integrable function of ${\omega }_2$ for almost all ${\omega }_1$.
• $f_1(x_1)={\int }_{{\Omega }_2}f(x_1,x_2)d{\mu }_2(x_2)$ is integrable almost surely on ${\Omega }_1$ ($f_1$ may be undefined or infinite on a zero measure set).
• $f_2(x_2)={\int }_{{\Omega }_1}f(x_1,x_2)d{\mu }_1(x_1)$ is integrable almost surely on ${\Omega }_2$.

And we have

$${\int }_{{\Omega }_1\times {\Omega }_2}fd\mu ={\int }_{{\Omega }_1}{f}_{1}d{\mu }_1={\int }_{{\Omega }_2}{f}_{2}d{\mu }_2.$$

Counterexamples

Non-Sigma-Finite Measure

Let ${\Omega }_1={\Omega }_2=(0,1)\subset \mathbb{R}$. Let ${\mathcal{F}}_1=\mathcal{B}(0,1)$, ${\mathcal{F}}_2=2^{{\Omega }_2}$. Let ${\mu }_1=\lambda$, the Lebesgue measure, and ${\mu }_2(A)=|A|$, the counting measure. Then ${\mu }_2$ is not $\sigma$-finite. Consider the function $f(x,y)=1\{x=y\}$. We have

$${\int }_{{\Omega }_1}{\int }_{{\Omega }_2}f(x,y)\mathrm{d}{\mu }_2\mathrm{d}{\mu }_1={\int }_{{\Omega }_1}1\mathrm{d}{\mu }_1=1,$$

but

$${\int }_{{\Omega }_2}{\int }_{{\Omega }_1}f(x,y)\mathrm{d}{\mu }_1\mathrm{d}{\mu }_2={\int }_{{\Omega }_2}0\mathrm{d}{\mu }_2=0,$$

General Function

Let ${\Omega }_1={\Omega }_2=\mathbb{N}$, ${\mathcal{F}}_1={\mathcal{F}}_2=2^{\mathbb{N}}$, and ${\mu }_1={\mu }_2$ be the counting measure. Consider the function $f(x,x)=1$, $f(x,x+1)=-1$, and $f(x,y)=0$ otherwise. Then

$${\int }_{{\Omega }_1}{\int }_{{\Omega }_2}f(x,y)\mathrm{d}{\mu }_2\mathrm{d}{\mu }_1=\sum _n\sum _{m}f(n,m)=\sum _{n}0=0,$$

but

$${\int }_{{\Omega }_2}{\int }_{{\Omega }_1}f(x,y)\mathrm{d}{\mu }_1\mathrm{d}{\mu }_2=\sum _m\sum _{n}f(n,m)=1+\sum _{m\ge 2}0=1.$$

Footnotes

1. A measure $\mu$ is $\sigma$-finite if there exists a countable collection of measurable sets $A_1,A_2,\dots$ such that $\mu (A_i)<\infty$ for all $i$ and $\Omega ={\bigcup }_i{A}_i$. ↩

2. That is $\int |f|\mathrm{d}\mu <\infty$. ↩