The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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July 28, 2010 - Posted by | Analysis, Measure Theory

2 Comments »

  1. […] sense that if either integral exists, then the other one does too, and their values are equal. As usual, it is sufficient to prove this for the case of for a measurable set . Linear combinations will […]

    Pingback by Pulling Back and Pushing Forward Structure « The Unapologetic Mathematician | August 2, 2010 | Reply

  2. […] can turn this into an iterated integral, which Fubini’s theorem tells us we can evaluate in any order we […]

    Pingback by Stokes’ Theorem (proof part 1) « The Unapologetic Mathematician | August 18, 2011 | Reply


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