A Quick Note on Fatou’s Lemma.
August 16, 2011
Above is a sweet picture of this French mathematician Pierre Fatou. Of course he had a nice mustache, but a few math things that are kind of neat about him: he was the first person to study the Mandelbrot set (though there is some controversy regarding the word "first" here…) and he also enjoyed iterative and recursive processes before they became really cool, making him a computer science hipster. Also, he has a lemma in measure theory named after him. That’s what this post is about.
Lemma (Fatou’s): Let be a sequence of non-negative measurable functions on with the Lebesgue measure , and let for all . Then, is measurable and we have the inequality
First, let me note that this actually works for any measure space, but to concrete it it suffices to think of the reals with the Lebesgue measure. Also, it may, in fact, be the case that these functions attain infinite values when we apply the limit; the inequality will still hold.
The proof of this is available in a ton of different places, but there’s usually an "easy way" (directly from the definitions) and a "hard way" (with monotone convergence theorem). I won’t show either.
But why is this called a "lemma"? As far as I can tell, this is one of the big pieces of something called the "Fatou-Lebesgue Theorem" which, given the same assumptions plus a measurable function such that for each , then
which is something we might suspect just from looking at Fatou’s lemma. And, in fact, proving this theorem is essentially just two applications of Fatou’s lemma.
Another place it is used is in the proof of the Dominated Convergence Theorem, which (despite being just a special case of the Fatou-Lebesgue Theorem) is used all over the place. I’ll most likely post about this one later if I find a good problem about it.
Probably the prototypical example of non-equality is the following:
Define if and let it be 0 elsewhere. We note immediately that the integral for each , and hence the limit infimum of the integrals is 1. On the other hand, we know that the limit of the will just be the zero function, which has an integral of 0. This agrees nicely with Fatou, since .
What does the what does the "inf" do in the statement? For some functions, we might have no nice limit. For example, let’s define . As , we don’t have any kind of nice limit. On the other hand, we do have a which is just the function . Note that the "right-hand side" of Fatou’s theorem is just going to be the lim inf of the measure of our space times . Of course, for the same reason as the left-hand-side, the lim inf of this will be -1 times the measure of the space. In this case the right and left-hand sides are equal, but if we remove the "inf" from the inequality, the statement becomes meaningless (what is the "limit" of as ?).
If anyone has any other neat applications of Fatou, lemme know. I mainly wrote this post up because I kept forgetting Fatou’s lemma, and I wanted to do some examples to drill it into my head. Last, I want to add that there is a Reverse Fatou’s Lemma, which is exactly what you think it is. Flip the inequality to and then replace "inf" with "sup." This, of course, is also used in the Fatou-Lebesgue Theorem. The same example as above now gives us the statement that . And that is reassuring.
Edit: AN ENTIRELY NEW SECTION!
Using Fatou to Show Completeness.
One of the things that I always find a bit intimidating is when questions ask me to show that some space is complete. As you know, this means that we need to show that every Cauchy sequence has a limit which is, itself, sitting in that space. It was not so obvious to me, for example, that a Cauchy sequence of functions ought to be . Fatou’s lemma helps with this kind of thing. Let’s just briefly state what we want to prove.
Definition. Let be an integer. A function is said to be in the space if , where our integral here is the Lebesgue integral.
For the purposes of this post, we will be using which are functions defined on the real line. It takes a little bit to prove that this is a metric space (in particular, we need to use Minkowski’s inequality to show the triangle inequality) with the metric given by
but after we have that we want to show that is, in fact, a complete metric space. Let’s just follow our noses here and see where we get to.
Proof (that is complete):
First, let’s fix a . Let’s stick with case, since the infinite case is done an easier way. Let’s first take a Cauchy sequence of functions. We’d like to show that this converges and that this converges to something in . This will test some of your understanding about modes of convergence as well. As ,
since the sequence is Cauchy, which implies
which implies that this sequence converges in measure. This implication is not direct, but essentially the argument is "How could fail to go to 0 when the integral does?" Since this converges in measure, there is a subsequence which converges almost uniformly which implies that this subsequences also converges almost everywhere. Let’s call the subsequence just for kicks. But, of course, this is a sequence of non-negative functions and so we may apply Fatou’s lemma! Note here that implies .
and we at last note that the right-hand side is finite, because these from above. Thus, we finally have:
which directly implies
which tells us that is in .