## Fubini’s Theorem

We continue our assumptions that and are both -finite measure spaces, and we consider the product space .

The first step towards the measure-theoretic version of Fubini’s theorem is a characterization of sets of measure zero. Given a subset , a necessary and sufficient condition for to have measure zero is that the -section have for almost all . Another one is that the -section have for almost all . Indeed, the definition of the product measure tells us that

Since the function is integrable and nonnegative, our condition for an integral to vanish says that the integral is zero if and only if -almost everywhere. Similarly, we see that the integral of is zero if and only if -almost everywhere.

Now if is a non-negative measurable function on , then we have the following equalities between the double integral and the two iterated integrals:

If is the characteristic function of a measurable set , then we find that

and thus

Next we assume that is a simple function. Then is a finite linear combination of characteristic functions of measurable sets. But clearly all parts of the asserted equalities are linear in the function , 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 , we can find an increasing sequence of simple functions converging pointwise to . The monotone convergence theorem tells us that

We define the functions

and conclude that since is an increasing sequence, must me an increasing sequence of non-negative measurable functions as well. For every the monotone convergence theorem tells us that

As a limit of a sequence of non-negative measurable functions, must also be a non-negative measurable function. One last invocation of the monotone convergence theorem tells us that

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 is an integrable function on , then almost every section of is integrable. If we define the functions

wherever these symbols are defined, then and are both integrable, and

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 . The latter equalities follow, then, from the above discussion. Since the measurable functions and 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 .

[…] 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 […]

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