## Nested Sequences of Measurable Sets.

### August 10, 2011

On nearly every practice qualifying exam that I’ve been studying from, the following question (in some guise) comes up:

**Question: **Let be a sequence of Lebesgue measurable sets in . Denote the Lebesgue measure by .

- If for every , then is it true that ? If not, add a criteria to make it true.
- If for every , then is it true that ? If not, add a criteria to make it true.

I’ll write up the solutions, since they are available elsewhere, but take a little bit of time to think about the second one if you haven’t. I’ll put the remainder of the post under a cut so that I don’t spoil it right away.

So, the second one is not true. If you haven’t thought of a counter-example, think about unbounded sets. You ought to get something from that. But still, we must prove that the first is true and that adding, "of finite measure" to the second will give us a true statement.

**Proof for 1: **Okay, first, we ought to note that the limit of these measurable sets is actually measurable. Otherwise, this doesn’t really make a whole lot of sense. This is done elsewhere, so I will not prove it. But the heart of the proof is in the following equalities. Let’s first just define to make things look nice. Then,

Do you see why this should be true? The sequences are nested in such a way that we are able to write the limit in terms of this union. But now notice that and are disjoint for all . This means that we’re just unioning up a lot of disjoint sets. By the definition of measure, we have disjoint additivity (also called *complete additivity*) and this allows us to write

which is nice so far. Okay, now, we’d like to end up with a limit on the outside of the measure (which, at this point, is inside the sum), so we note that taking an infinite sum is like taking a limit in the following way:

but, of course, this sum is just adding up these disjoint pieces, and these disjoint pieces union up to (check this!):

This is exactly what we wanted. This proves that **1** is true with no modifications to the statement required. If you want a slick proof, just read the equations and ignore the paragraphs.

**Proof for 2: ** This one is a bit trickier, since we first note that it is not true (you provided a counter-example!) but that the only problem seems to be that we need things that have finite measure.

The crux of the proof is the following little property: if and , then . This is not hard to see, since and are disjoint, so we can just add up their measures. But when is of infinite measure, then we get into a sticky situation: take, for example, and . Clearly, the difference has measure 1, but the proposition says that the measure should be which is not defined. We require, therefore, to be of finite measure which forced to have finite measure.

Did that example seem familiar? It’s virtually the same one that works as a counter-example to the second statement. Thus, we require *at least one (and hence infinitely many) of the to be of finite measure.* Let’s call the first one that has finite measure . We then evaluate this in two different ways. The first one will use the property that we just stated above.

.

But since is (eventually) a nested *increasing* sequence, we use the first problem’s conclusion (!) to justify this:

.

Subtracting one from the other, we get exactly the statement in part **2**, but we’ve needed to use the fact that at least one (and hence infinitely many) of the sets have finite measure. Bam.

Nice. Can you give an example of a sequence of functions that converge to 0 in measure but not in L^1? How about vice versa? Can you give a sequence of functions that converges nowhere pointwise but converges in L^1 to 0?

Prove that the limit of ||f||_{L^p} as p \rightarrow \infty is ||f||_{L^{\infty}} (if f belongs to this space).

Wonderful read, I recently handed down this upon an associate who had previously been conducting a very little study for that will. And he in fact bought us lunch since I recently found that for him smile Therefore let me rephrase of which: Many thanks for break!