Today we prove Fatou’s Lemma, which is a precursor to the Fatou-Lebesgue theorem, and an important result in its own right.
If is a sequence of non-negative integrable functions then the function defined pointwise as
is also integrable, and we have the inequality
In fact, the lemma is often stated for a sequence of measurable functions and concludes that is measurable (along with the inequality), but we already know that the limit inferior of a sequence of measurable functions is measurable, and so the integrable case is the most interesting part for us.
So, we define the functions
so that each is integrable, each and the sequence is pointwise increasing. Monotonicity tells us that for each we have
and it follows that
We also know that
which means we can bring the monotone convergence theorem to bear. This tells us that
If it happened that were not integrable, then some of the would have to be only measurable — not integrable — themselves. And it couldn’t just be a finite number of them, or we could just drop them from the sequence. No, there would have to be an infinite subsequence of non-integrable , which would mean an infinite subsequence of their integrals would diverge to . Thus when we take the limit inferior of the integrals we get , as we do for the integral of itself, and the inequality still holds.
We want to prove a strengthening of the dominated convergence theorem. If is an a.e. increasing sequence of extended real-valued, non-negative, measurable functions, and if converges to pointwise a.e., then
If is integrable, then dominates the sequence , and so the dominated convergence theorem itself gives us the result we assert. What we have to show is that if , then the limit diverges to infinity. Or, contrapositively, if the limit doesn’t diverge then must be integrable.
But this is the limit of a sequence of real numbers, and so if it converges then it’s Cauchy. That is, we can conclude that
Our assumption that is a.e. increasing tells us that for any fixed and , the difference is either a.e. non-negative or a.e. non-positive. That is,
And thus the sequence is mean Cauchy, and thus mean convergent to some integrable function , which must be equal to almost everywhere.
One nice use of this is when talking about series of functions. If is a sequence of integrable functions so that
then I say that the series
converges a.e. to an integrable function , and further that
To see this, we let be the partial sum
which gives us a pointwise increasing sequence of non-negative measurable functions. The monotone convergence theorem tells us that these partial sums converge pointwise to some and that
But this is exactly the sum we assumed to converge before. Thus the function is integrable and the series of the is absolutely convergent. That is, since must be a.e. finite, the series
is absolutely convergent for almost all , and so it must be convergent pointwise almost everywhere. Since dominates the partial sums
the bounded convergence theorem tells us that limits commute with integrations here, and thus that
If is an essentially bounded measurable function with a.e. for some real numbers and , and if is any integrable function, then there is some real number with so that
Actually, this is a statement about finite measure spaces; the function is here so that the indefinite integral of will give us a finite measure on the measurable space to replace the (possibly non-finite) measure . This explains the in the multivariable case, which wasn’t necessary when we were just integrating over a finite interval in the one-variable case.
Okay, we know that a.e., and so a.e. as well. This tells us that is integrable. And thus we conclude
Now either the integral of is zero or it’s not. If it’s zero, then is zero a.e., and so is , and our assertion follows for any we like. On the other hand, if it’s not we can divide through to find
this term in the middle is our .
Now that we’ve established Lebesgue’s dominated convergence theorem, we can put it to good use.
If is a measurable function and is an integrable function so that a.e., then is integrable. Indeed, we can break any function into positive and negative parts and , which themselves must satisfy a.e., and which are both nonnegative. So if we can establish the proposition for nonnegative functions the general case will follow.
If is simple, then it has to be integrable or else couldn’t be. In general, there is an increasing sequence of nonnegative simple functions converging pointwise to . Each of the is itself less than , and thus is integrable; we find ourselves with a sequence of integrable functions, dominated by the integrable function , converging pointwise to . The dominated convergence theorem then tells us that is integrable.
Next, a measurable function is integrable if and only if its absolute value is integrable. In fact, we already know that being integrable implies that is, but we can now go the other way. But then is a measurable function and an integrable function with everywhere. The previous result implies that is integrable.
If is integrable and is an essentially bounded measurable function, then the product is integrable. Indeed, if a.e., then a.e. as well. Since is integrable, our first result tells us that is integrable, and our second result then tells us that is integrable.
Finally, for today, if is an essentially bounded measurable function and is a measurable set of finite measure, then is integrable over . Since has finite measure, its characteristic function is integrable. Then our previous result tells us that is integrable, which is what it means for to be integrable over .
The dominated convergence theorem provides a nice tool to make sure certain sequences of integrable functions converge (in the mean) to integrable limits. Yes, we have the definition and the characterization in terms of convergence in measure, but this theorem is often easier to apply.
If is a sequence of integrable functions converging in measure or converging a.e. to a function , and if is an integrable function which “dominates” the sequence — we have for almost all — then is integrable and the sequence converges to in the mean. It’s important to note that we do not assume that is integrable; that’s part of the conclusion.
If we start by assuming that converges in measure to , then yesterday’s result immediately tells us that converges in the mean to ; the uniformity assumptions from that theorem are consequences of the inequalities
Now, if we only assume that converges a.e. to , then we can reduce to convergence in measure by using . By throwing out a set of measure zero, we will assume that and are all no more than everywhere. Then for every fixed positive we can write
but the measure of this latter set is finite, and so for all . A.e. convergence tells us that the measure of the intersection of all the is . By continuity, we conclude that
That is, in the presence of a dominating function , convergence a.e. implies convergence in measure, and thus implies convergence in mean.
Incidentally, since we know that
we can use convergence in the mean to control the right hand side, and thus get control over the left hand side. That is, we find that
The dominated convergence theorem shows that the integral and the limit commute so long as the sequence is dominated by some integrable function.
It should be noted that the dominating function is essential. Indeed, let be the closed unit interval with Lebesgue measure, and let . Now we consider the sequence which converges in measure to . However, there is no integrable dominating function; we find that
and so the sequence cannot converge in the mean to .
First off we want to introduce another notion of continuity for set functions. We recall that a set function on a class is continuous from above at if for every decreasing sequence of sets with we have . If is a sequence of set functions, then, we say the sequence is “equicontinuous from above at ” if for every sequence decreasing to and for every there is some number so that if we have . It seems to me, at least, that this could also be called “uniformly continuous from above at “, but I suppose equicontinuous is standard.
Anyway, now we can characterize exactly how convergence in mean and measure differ from each other: a sequence of integrable functions converges in the mean to an integrable function if and only if converges in measure to and the indefinite integrals of are uniformly absolutely continuous and equicontinuous from above at .
We’ve already shown that convergence in mean implies convergence in measure, and we’ve shown that convergence in mean implies uniform absolute continuity of the indefinite integrals. All we need to show in the first direction is that if converges in mean to , then the indefinite integrals are equicontinuous from above at .
For every we can find an so that for we have . The indefinite integral of a nonnegative a.e. function is real-valued, countably additive, and nonnegative, and thus is a measure. Thus, like any measure, it’s continuous from above at . And so for every sequence of measurable sets decreasing to there is some so that for we find
the first for all from to . Then if we have
for every positive . We control the first term in the middle by the mean convergence of for and by the continuity from above of for . And so the are equicontinuous from above at .
Now we turn to the sufficiency of the conditions: assume that converges in measure to , and that the sequence of indefinite integrals is both uniformly absolutely continuous and equicontinuous from above at . We will show that converges in mean to .
We’ve shown that is -finite, and so the countable union
of all the points where any of the are nonzero is again -finite. If is an increasing sequence of measurable sets with , then the differences form a decreasing sequence converging to . Equicontinuity then implies that for every there is some so that , and thus
For any fixed we define
and it follows that
By convergence in measure and uniform absolute continuity we can make the integral over arbitrarily small by choosing and sufficiently large. We deduce that
and, since was arbitrary, we conclude
Now we can see that
and, since is arbitrary,
That is, the sequence is Cauchy in the mean. But we know that the norm is complete, and so converges in the mean to some function . But this convergence in mean implied convergence in measure, and so converges in measure to , and thus almost everywhere.
I evidently have to avoid saying in post titles because WordPress’ pingbacks can’t handle the unicode character ¹…
Anyhow, today we can show that the norm gives us a complete metric structure on the space of all integrable functions on . But first:
If is a mean Cauchy sequence of integrable simple functions converging in measure to , then it converges in the mean to as well. Indeed, for any fixed the sequence is a mean Cauchy sequence of integrable simple functions converging in measure to . Thus we find
But the fact that is mean Cauchy means that the integral on the right gets arbitrarily small as we take large enough and . And so on the left, the integral gets arbitrarily small as we take a large enough . That is,
and converges in the mean to .
As a quick corollary, given any integrable function and positive number , there is some integrable simple function with . Indeed, since is integrable there must be some sequence of integrable simple functions converging in measure to it, and we can pick for some sufficiently large .
Now, if is a mean Cauchy sequence of integrable functions — any integrable functions — then there is some integrable function to which the sequence converges in the mean. The previous corollary tells us that for every there’s some integrable simple so that . This gives us a new sequence , which is itself mean Cauchy. Indeed, for any choose an large enough so that . Then for we have
Since is Cauchy in measure it converges in measure to some function . Then our first result shows that also converges in the mean to — that goes to zero as goes to . But we can also write
so if we choose a large enough , this will get arbitrarily small as well. That is, also converges in the mean to .
This shows that any mean Cauchy sequence of integrable functions is also mean convergent to some function, and thus the space of integrable functions equipped with the norm is a Banach space.
We’ve got some interesting results about when integrals come out to be zero.
First up: if is an a.e. nonnegative integrable function, then if and only if almost everywhere. In one direction, if a.e. then we can pick all as a sequence of simple functions converging in measure to ; clearly the limit of their integrals is zero. On the other hand if a mean Cauchy sequence of integrable simple functions converging in measure to , then so does since is a.e. nonnegative. If then we must have
that is, converges in mean to , which we know means it converges in measure to as well. But we picked this sequence to converge in measure to , and so almost everywhere.
As a quick corollary, if is integrable and is a set of measure zero, then , because this is really the integral of , which is a.e. zero.
A sort of a converse: if is integrable and positive a.e. on a measurable set , and if , then . We define and
for all positive integers ; our assumption tells us that has measure zero, so if we can show that does too, then we’ll see that . But we have
where the last inequality holds because a.e. on . This shows that for all . But is the union of all the , and so
as we wanted to show.
Now if is integrable and for every measurable , then almost everywhere. Indeed, if , then our hypothesis says that , and the previous result then shows that . Applying the same reasoning to shows that the same is true of the set of points where is negative.
Finally, we can show that if is integrable, then the set is -finite. To see this, let be a mean Cauchy sequence of integrable simple functions converging in measure to . For every , is a measurable set of finite measure. Now set
If is any measurable subset of , then we can see that
since is outside each . Now our previous result shows us that a.e. on , but since we have everywhere in ! The only possibility is that itself has measure zero. And thus
writes as the (countable) union of a bunch of sets of finite measure, as desired.
First off, the basic definitions of the norm carries across: we set
and we say that a sequence of integrable functions is Cauchy in the mean or is mean Cauchy if goes to zero for sufficiently large and . We immediately find that any sequence that is Cauchy in the mean is also Cauchy in measure, since our earlier proof didn’t really depend on the functions being simple.
We also have a couple more results whose proofs don’t really depend on the simplicity of , and so these carry across without change. If is mean Cauchy, with indefinite integrals , then the collection of set functions is uniformly absolutely continuous. Further, we can define the limiting set function
for every measurable , and we find that this set function is finite-valued and countably additive.
So now we can formally define the obvious notion: the sequence of integrable functions “converges in the mean” or is “mean convergent” to an integrable function if goes to zero as goes to . And, as we might expect, if converges in the mean to then it converges in measure as well.
Indeed, for every we define
and then we see that
Thus must go to as , and so converges in measure to .
Now we’ve got a definition of the integral of a wider variety of functions than before, so let’s look over the basic properties.
First of all, from what we know about convergence in measure and algebraic and order properties of integrals of simple functions, we can see that if and are integrable functions and is a real number, then so are the absolute value , the scalar multiple , and the sum . As special cases, we can see that the positive and negative parts
are both integrable.
If is a measurable set and is a mean Cauchy sequence of integrable simple functions converging in measure to , then it should be clear that is mean Cauchy and converges in measure to . Thus we can define
Since integrals of simple functions are linear, and limits are linear, we immediately conclude that
as before, but now for general integrable functions. Similarly, if is nonnegative a.e., we can find a sequence of nonnegative simple functions converging a.e. (and thus in measure) to . The integral of each of these functions is nonnegative, and so their limit must be as well. That is, if a.e., then
Now, all our properties that we proved using only these two linearity and order properties follow. If a.e., then
For any two integrable functions amd we find
If and are real numbers so that for almost all , then we have
and if an integrable function is a.e. nonnegative, then its indefinite integral is monotone.
It takes a bit of work, though, to check that the indefinite integral of an integrable function is absolutely continuous. If is a mean Cauchy sequence converging in measure to , then we have
We know that the indefinite integrals of the are uniformly absolutely continuous, so we have control over the size of the first term on the right. The second term on the right can be kept small by choosing a large enough , since converges in measure to .
Finally, the indefinite integral of is countably additive; since is a mean Cauchy sequence of simple functions we can write as the limit , and we know that this limit is countably additive.