We skipped this in my analysis class, so I’m going to prove Fubini’s theorem in this post. I’m following the proof from Stein and Shakarchi’s Real Analysis with some restructuring. The basic result is that you can compute integrals in by first integrating in , then in , and the order in which you do these two integrals doesn’t matter.
To be precise, we can think of points in as , where and . Then, for each , we can define the “slices” of a function and a set corresponding to as
(Of course, we could define slices corresponding to as well, but it’s sufficient to do everything for -slices, since the proofs for -slices are completely analogous.)
Theorem 1 (Fubini’s Theorem). Let be the set of functions on satisfying:
- The slice is integrable on for almost every
- The function is integrable on
Then i.e. every integrable function is in .
The idea of the proof is to show that contains the simple functions, and then approximate functions by simple functions. This approach requires two facts: that is closed under linear combinations, and that is closed under monotone limits. We start with these two proofs.
Lemma 2. A finite linear combination of functions in is in .
Proof: Suppose we have . Then for each , is integrable on except for , where . If we let , then , and if , then will be integrable except when . It follows that satisfies 1 and 2, and then the linearity of the integral gives 3, so .
Lemma 3. A monotone limit of functions in is in .
Proof: Suppose we have . We can make two simplifying assumptions: that the convergence is monotonically increasing, and that the are non-negative. If either of these is not the case, then following argument can be applied to and to give the result.
The monotone convergence theorem tells us that
We know that for every , is integrable except for where , and if we let , then , and is integrable on for all except for . Using the monotone convergence theorem again gives
A third application of monotone convergence gives that
Now, because , we know that
and this together with our previous work gives that
The RHS is finite because is integrable, so is integrable as well. Thus almost everywhere, and so is integrable for almost every , and
It follows that .
Next, we show that contains the simple functions. The idea here is to start with the characteristic functions of sets and null sets, and then use our previous work to get to simple functions.
Lemma 4. If is a simple function, then .
Proof: We build this up in steps.
- If is a set of finite measure, then .
- If is a bounded open cube, then .
Let be such an open cube, where are cubes in and respectively. It’s clear that satisfies 1 and 2, so we need to do the computation to verify 3.
For each , the function is measurable in and integrable with
Thus is also measurable and integrable, and
But of course,
- If is a subset of the boundary of a closed cube, then .
Let this set be . It’s still easy to check 1 and 2, so we verify 3. Because the boundary of the cube has measure zero in , we always have
On the other hand, for almost every , the slice has measure zero in , and so the iterated integral is also zero. It follows that .
- If is the finite union of closed cubes with disjoint interior, then .
The idea is to write , where each is either the interior of an open cube, or a subset of the boundary of a closed cube. From 1.1 and 1.2, each , and now Lemma 2 gives that .
- If is open and of finite measure, then .
Such an open set can be written as the countable union of almost disjoint (i.e. their intersection is null) closed cubes , so we define
Then, , which is integrable because has finite measure, and Lemma 3 allows us to conclude that .
- If is and of finite measure, then .
Because is , there are open sets such that
and because is measurable there is an open set . Then, let
and we have a decreasing sequence of open sets
Now, and by 1.4, , so Lemma 3 shows that .
- If is a bounded open cube, then .
- If , then .
Because a null set is measurable, we can find a set such that . From Step 1, we know that , so
for almost every . This means that and , so we have that for almost every , and so
- If is measurable with finite measure, then .
Recall that we can find a set such that , and now writing and Lemma 2 gives that .
Now, any simple function is a linear combination of characteristic functions of measurable sets with finite measure, so Lemma 2 shows that .
With these three lemmas in hand, we can prove Fubini’s theorem.
Proof of Fubini’s Theorem: Suppose is an integrable function. We can write as the sum of a positive and negative part, so it is sufficient by Lemma 2 to consider the case where is non-negative. Because is integrable, there are simple functions that converge monotonically to . By Lemma 4, each , and now Lemma 3 allows us to conclude that .