# The Unapologetic Mathematician

## Fubini’s Theorem

We continue our assumptions that $(X,\mathcal{S},\mu)$ and $(Y,\mathcal{T},\nu)$ are both $\sigma$-finite measure spaces, and we consider the product space $(X\times Y,\mathcal{S}\times\mathcal{T},\mu\times\nu)$.

The first step towards the measure-theoretic version of Fubini’s theorem is a characterization of sets of measure zero. Given a subset $E\subseteq X\times Y$, a necessary and sufficient condition for $E$ to have measure zero is that the $X$-section $E_x$ have $\nu(E_x)=0$ for almost all $x\in X$. Another one is that the $Y$-section $E^y$ have $\mu(E^y)=0$ for almost all $y\in Y$. Indeed, the definition of the product measure tells us that $\displaystyle\lambda(E)=\int\nu(E_x)\,d\mu(x)=\int\mu(E^y)\,d\nu(y)$

Since the function $x\mapsto\nu(E_x)$ is integrable and nonnegative, our condition for an integral to vanish says that the integral is zero if and only if $\nu(E_x)=0$ $\mu$-almost everywhere. Similarly, we see that the integral of $\mu(E^y)$ is zero if and only if $\mu(E^y)=0$ $\nu$-almost everywhere.

Now if $h$ is a non-negative measurable function on $X\times Y$, then we have the following equalities between the double integral and the two iterated integrals: $\displaystyle\int h\,d(\mu\times\nu)=\iint h\,d\mu\,d\nu=\iint h\,d\nu\,d\mu$

If $h$ is the characteristic function $\chi_E$ of a measurable set $E$, then we find that \displaystyle\begin{aligned}\int\chi_E(x,y)\,d\nu(y)&=\int\chi_{E_x}(y)\,d\nu(y)=\nu(E_x)\\\int\chi_E(x,y)\,d\mu(x)&=\int\chi_{E^y}(x)\,d\mu(x)=\mu(E^y)\end{aligned}

and thus \displaystyle\begin{aligned}\iint\chi_E(x,y)\,d\nu\,d\mu=\int\nu(E_x)\,d\mu&=\left[\mu\times\nu\right](E)=\int\chi_E\,d(\mu\times\nu)\\\iint\chi_E(x,y)\,d\mu\,d\nu=\int\mu(E^y)\,d\nu&=\left[\mu\times\nu\right](E)=\int\chi_E\,d(\mu\times\nu)\end{aligned}

Next we assume that $h$ is a simple function. Then $h$ is a finite linear combination of characteristic functions of measurable sets. But clearly all parts of the asserted equalities are linear in the function $h$, and so since they hold for characteristic functions of measurable sets they must hold for any simple function as well.

Finally, given any non-negative measurable function $h$, we can find an increasing sequence of simple functions $\{h_n\}$ converging pointwise to $h$. The monotone convergence theorem tells us that $\displaystyle\lim\limits_{n\to\infty}\int h_n\,d(\mu\times\nu)=\int h\,d(\mu\times\nu)$

We define the functions $\displaystyle f_n(x)=\int h_n(x,y)\,d\nu(y)$

and conclude that since $\{h_n\}$ is an increasing sequence, $\{f_n\}$ must me an increasing sequence of non-negative measurable functions as well. For every $x$ the monotone convergence theorem tells us that $\displaystyle\lim\limits_{n\to\infty}f_n(x)=f(x)=\int h(x,y)\,d\nu(y)$

As a limit of a sequence of non-negative measurable functions, $f$ must also be a non-negative measurable function. One last invocation of the monotone convergence theorem tells us that $\displaystyle\lim\limits_{n\to\infty}\int f_n\,d\mu=\int f\,d\mu$

which proves the equality of the double integral and one of the iterated integrals. The other equality follows similarly.

And now we come to Fubini’s theorem itself: if $h$ is an integrable function on $X\times Y$, then almost every section of $h$ is integrable. If we define the functions \displaystyle\begin{aligned}f(x)&=\int h(x,y)\,d\nu(y)\\g(y)&=\int h(x,y)\,d\mu(x)\end{aligned}

wherever these symbols are defined, then $f$ and $g$ are both integrable, and $\displaystyle\int h\,d(\mu\times\nu)=\int f\,d\mu=\int g\,d\nu$

Since a real-valued function is integrable if and only if both its positive and negative parts are, it suffices to consider non-negative functions $h$. The latter equalities follow, then, from the above discussion. Since the measurable functions $f$ and $g$ have finite integrals, they must be integrable. And since they’re integrable, they must be finite-valued a.e., which implies the assertions about the integrability of sections of $h$.