Real Analysis Primer, part 3: Lebesgue Measure and a Non-measurable Set.
May 28, 2011
Last time, and some Equivalent Statements.
Last time we talked about the Lebesgue measure in terms of outer and inner measure. Recall that outer measure of a set is this "almost-measure" that we made up by taking the infimum of sets that looked like disjoint intervals which covered the set we’re measuring. Recall that we worked in the unit interval and recall that is an elementary set if it is a countable union of intervals (half-open, open, or closed). Then we can better write outer measure as
and we also had a corresponding "inner measure" which was another almost-measure, defined in the following way:
and we pieced these together to make an actual measure, called the Lebesgue measure, which is defined on every set where the inner and outer measures match up. In other words, if , then exists and .
Let’s just note something here. We really didn’t need to define inner measure at all. Note that
which, when we rewrite this, expresses that is Lebesgue measurable if
And this is perfectly reasonable. Intuitively, this reads, "If the outer measure of plus the outer measure of the compliment of sum up to the measure of the unit interval, then is measurable."
It might still be unsettling to you that there are sets which are non-measurable, but let’s show that this is indeed the case. Before we do this, let’s note two propositions which are interesting in their own right.
Proposition. Outer measure is translation invariant. If and , then .
This is relatively obvious, since we’re not changing the size of , but it may be the case that is not actually in anymore. Because we could easily extend the domain in which we are working, I won’t make too much of a fuss about this. Either way, the idea of the proof is to note that intervals (and therefore elementary sets) are translation invariant, and then just cover the translated set with translated elementary sets.
One thing that I want to mention before going into the second property is a notion of closeness. We want to consider arbitrary sets , but we know very little about . We know quite a bit about elementary sets, though, and so we can get "really close" to by covering it with elementary sets that estimate it better and better. In fact, because we defined the outer measure as an infimum, for any positive number we have some elementary set covering such that . This means that no matter how close we get to the actual measurement of , there will be some elementary set that we can use to approximate that "close measure." We will use this idea a lot. The next proof is a great example of how to apply this.
Proposition. Outer measure satisfies what is called "countable subadditivity", which means that for any countable collection of subsets .
Proof. We first let . We know nothing about each since they’re arbitrary sets. But, as above, we can estimate them in a nice way. The following trick is used nearly an uncountable number of times in analysis, so pay attention. We have a countable number of the so we’ll estimate them in the following way:
For each there exists an elementary set covering such that . Now we note that
Where the last inequality is because some of the may overlap. We then note that from above, we have
Oh! So that’s what that was there. Yes, the fact that is something which comes up constantly; especially when we need to add a countable number of things together and get an epsilon to pop out. So, all together, what does this inequality say? Since this is true for all , it follows that
which is exactly what we wanted to show.
An Unwelcome Guest.
We now have enough to show that there exists a set which is not Lebesgue measurable. Instead of working in , we need to work in something slightly bigger for the numbers to work out nicely. For the time being, let’s pretend we’re working in , but since all of our sets will turn out to be bounded, there will be no problems coming up with infinite measures. I’ll just sketch the idea since the actual proof is available a number of places. The set in question is so famous that it even has a name: it’s called the Vitali set. The same proof cam be applied, though, to show that there are actually an uncountable number of non-measurable sets.
Proposition. There is a subset of which is not Lebesgue measurable.
Proof. We need to start by considering a quotient group. For those of you who have not taken algebra, I’ll just write this out explicitly. Consider the set
This is a set of translates of . So, for example, one element of this set is which is, itself, a subset of consisting of for every . So, every element of our set above is a subset which is actually a translate of the rational numbers.
Convince yourself that each of these subsets (the translates of ) each have at least one element in them which is in the subset . So, for example, in , the element . Similarly, each of these translates contains some (in fact, many) element in . Using the Axiom of Choice we pick out ONE element from each of these translates of that is in and put them into a set . So, our set . A fancy way to say this (using algebra) is our set contains one element from each coset of .
Now we’ve constructed and all we need to do is show that it is non-measurable. The way that we’ll do this is by supposing it is, and then deriving a contradiction. We’ll break this into steps so that it’s easier to follow.
We create translations of in the following way. Let be an enumeration of the rationals between [-1,1] (since they’re countable, we can enumerate them). Now define . This is simply a translate of , no big deal.
We show that . Suppose . Then by our construction of , we considered and took one of the elements that was in in this subset; call that element . Then and so their difference is a rational. Since they’re both in , the smallest their difference can be is -1 and the largest is can be is 1; hence there is some such that . But this implies that . Good. This shows what we wanted. This might be a bit confusing to read, but just write it down and you’ll see what I mean.
We now show that . This is nearly trivial; no element can exceed 1 + 1 = 2, and no element can be less than 0 + (-1) = -1.
We now note that since , this implies that the Lebesgue measure of the union of the is greater than or equal to the measure of and less than or equal to the measure of . In other words;
but now here’s the kicker. We proved right before this that Lebesgue measurable sets are translation invariant. But then . Clever, clever. So, what this really says is:
but this is no good. Suppose . Then summing it up infinitely many times gives a divergent sequence. Similarly if . If , then the sum will be 0, which is not between 1 and 3. We have arrived at a contradiction. We must have that is not measurable.
This set is hard to imagine. You can try; it’s not very fun. Note that a crucial part of this proof used the axiom of choice which makes strange things happen often; thus, you should tell yourself that even though there are non-measurable sets out there, we needed to use a something as powerful as the AoC to find one. So Lebesgue measure is pretty good, all things considered.
In addition to non-measurable sets, the other kinds of sets which come up quite often are sets of measure zero. They have a lot of special properties, and sort of function as sets which the Lebesgue measure "doesn’t see" or "doesn’t care about." Next time we’ll go into this a bit more.