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 [0,1] and recall that E\subseteq [0,1] 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

\displaystyle \mu^{\ast}(A) = \inf_{A\subseteq E}\{\mu^{\ast}(E)\}

and we also had a corresponding "inner measure" which was another almost-measure, defined in the following way:

\mu_{\ast}(A) = 1 - \mu^{\ast}([0,1]-A)

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 \mu^{\ast}(A) = \mu_{\ast}(A), then \mu(A) exists and \mu(A) = \mu^{\ast}(A)

Let’s just note something here.  We really didn’t need to define inner measure at all.  Note that

\mu^{\ast}(A) = \mu_{\ast}(A) = 1 - \mu^{\ast}([0,1]-A)

which, when we rewrite this, expresses that A is Lebesgue measurable if

\mu^{\ast}(A) + \mu^{\ast}([0,1]-A) = 1

And this is perfectly reasonable.  Intuitively, this reads, "If the outer measure of A plus the outer measure of the compliment of A sum up to the measure of the unit interval, then A 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 r\in {\mathbb R} and A\subseteq [0,1], then \mu^{\ast}(A + r) = \mu^{\ast}(A).


This is relatively obvious, since we’re not changing the size of A, but it may be the case that A + r is not actually in [0,1] 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 A, but we know very little about A.  We know quite a bit about elementary sets, though, and so we can get "really close" to A 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 \epsilon > 0 we have some elementary set E covering A such that \mu^{\ast}(E) < \mu^{\ast}(A) + r.  This means that no matter how close we get to the actual measurement of A, 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 \displaystyle \mu^{\ast}\left(\bigcup_{i}(A_{i})\right)\leq \sum_{i}\mu^{\ast}(A_{i}) for any countable collection of subsets A_{i}\subseteq [0,1]


Proof.  We first let \epsilon > 0.  We know nothing about each A_{i} 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 A_{i} so we’ll estimate them in the following way:

For each A_{k} there exists an elementary set E_{k} covering A_{k} such that \mu^{\ast}(E_{k}) < \mu^{\ast}(A_{k}) + \frac{\epsilon}{2^{k}}.  Now we note that

\displaystyle \mu^{\ast}\left(\bigcup_{i}(A_{i})\right)\leq \mu^{\ast}\left(\bigcup_{i}(E_{i})\right) \leq \sum_{i}\mu^{\ast}(E_{i})

Where the last inequality is because some of the E_{i} may overlap.  We then note that from above, we have

\displaystyle \sum_{i}\mu^{\ast}(E_{i}) < \sum_{i}\mu^{\ast}(A_{i}) + \frac{\epsilon}{2^{i}} = \sum_{i}\mu^{\ast}(A_{i}) + \sum_{i}\frac{\epsilon}{2^{i}}

\displaystyle = \sum_{i}\mu^{\ast}(A_{i}) + \epsilon

Oh!  So that’s what that 2^{k} was there.  Yes, the fact that \sum_{i=1}^{\infty} \frac{1}{2^{i}} = 1 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 \epsilon > 0, it follows that

\displaystyle \mu^{\ast}\left(\bigcup_{i}(A_{i})\right) \leq \sum_{i}\mu^{\ast}(A_{i})

which is exactly what we wanted to show.  \Box


An Unwelcome Guest.

We now have enough to show that there exists a set which is not Lebesgue measurable.  Instead of working in [0,1], 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 {\mathbb R}, 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 [0,1] 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

\{r + {\mathbb Q}\, |\, r\in {\mathbb R}\}.

This is a set of translates of {\mathbb Q}.  So, for example, one element of this set is \pi + {\mathbb Q} which is, itself, a subset of {\mathbb R} consisting of \pi + q for every q\in {\mathbb Q}.  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 {\mathbb Q}) each have at least one element in them which is in the subset [0,1].  So, for example, in \pi + {\mathbb Q}, the element \pi + (-\frac{22}{7}) \approx 0.00126\dots\in [0,1].  Similarly, each of these translates contains some (in fact, many) element in [0,1].  Using the Axiom of Choice we pick out ONE element from each of these translates of {\mathbb Q} that is in [0,1] and put them into a set V.  So, our set V\subseteq [0,1].  A fancy way to say this (using algebra) is our set V contains one element from each coset of {\mathbb R}/{\mathbb Q}

Now we’ve constructed V 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.

Step 1. 

We create translations of V in the following way.  Let \{q_{i}\} be an enumeration of the rationals between [-1,1] (since they’re countable, we can enumerate them).  Now define V_{i} = V + q_{i}.  This is simply a translate of V, no big deal. 

Step 2.

We show that [0,1]\subseteq \bigcup_{i}V_{i}.  Suppose x\in [0,1].  Then by our construction of V, we considered x + {\mathbb Q} and took one of the elements that was in [0,1] in this subset; call that element y.  Then x - y \in {\mathbb Q} and so their difference is a rational.  Since they’re both in [0,1], the smallest their difference can be is -1 and the largest is can be is 1; hence there is some q_{k}\in [-1,1] such that x - y = q.  But this implies that x\in V_{k}.  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.

Step 3.

We now show that \bigcup_{i}V_{i}\subseteq [-1,2].  This is nearly trivial; no element can exceed 1 + 1 = 2, and no element can be less than 0 + (-1) = -1.

Step 4.

We now note that since [0,1]\subseteq \bigcup_{i}V_{i}\subseteq [-1,2], this implies that the Lebesgue measure of the union of the V_{i} is greater than or equal to the measure of [0,1] and less than or equal to the measure of [-1,2].  In other words;

1 \leq \sum_{i=1}^{\infty}\mu(V_{i}) \leq 3

but now here’s the kicker.  We proved right before this that Lebesgue measurable sets are translation invariant.  But then \mu(V_{i}) = \mu(V).  Clever, clever.  So, what this really says is:

1 \leq \sum_{i=1}^{\infty}\mu(V) \leq 3

but this is no good.  Suppose \mu(V) > 0.  Then summing it up infinitely many times gives a divergent sequence.  Similarly if \mu(V) < 0.  If \mu(V) = 0, then the sum will be 0, which is not between 1 and 3.  We have arrived at a contradiction.  We must have that V is not measurable.  \Box


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.


Next Time.

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.

12 Responses to “Real Analysis Primer, part 3: Lebesgue Measure and a Non-measurable Set.”

  1. Joe said

    Great blog! It’s helping me understand ideas in RA. You have a knack for explanation.

    I came across a question in which the answer I don’t understand, and since it’s related to this topic, I thought I’d throw it to you.

    “Give an example of a disjoint sequence of sets Ej in R such that m*(U Ej) < ∑m*(Ej)."

    It seems the answer relies on the fact that you must choose a non-measurable set A first. So, in [0, 1], choose A non-measurable. Then we'd know that m*(A) = m*[(A+r)(mod 1)], r in Q. Then, m*U[(A+r)(mod 1)] = [0,1] = 1. But the sum of the measures is infinity since the measure of A is positive. Clever.

    A couple of questions: The mod 1 is confusing. And, why couldn't we have chosen 0 < A < ∞ and measurable?

    • James said

      Thank you! It’s been a bit since I’ve done this, but I’ll attempt to motivate this. If your sets were measurable, then the measure of the disjoint union would be equal to the sum of the measures of each piece — this is true even in this case that they are “almost” disjoint, but in your case it’s nicer. The proof of this is probably in any standard text, but it “seems” reasonable. So, the only way that you can have a disjoint sequence of sets that have such a strict inequality going on would be if they were *not* measurable. Hence, we must look for a nonmeasurable set.

      As it turns out, there is a standard nonmeasurable set that everyone loves to use. In fact, I construct it above! It satisfies the criteria nicely.

      As for your other part: suppose that we chose A measurable such that the measure of A was strictly positive and finite; then, since the measure is translation invariant, translating it a countable number of times would give us countably many positive finite-measured sets; the sum of the measures would be infinite and, similarly, the union of the translations would similarly be infinite (I think this is just subjectivity), which gives an equality \infty = \infty.

  2. Joe said

    Thanks, James. Yes, I see now. If A were measurable, we’d have an infinite measure for the sum of outer measures.

    A quick question on your step 1 from above. You mention that the Vi’s are a translate of the V’s, so the measure of Vi is the measure of V. But q is [-1, 1] so wouldn’t that expand V to [-1, 1]?

    • James said

      I’m not entirely sure what you mean; but each q_{i} is just some rational number in [-1,1]. Regardless of what V is, forming the V + q_{i}‘s just kind of gives us countably many copies of V which are translated over a little bit. This does expand the whole set (translates, etc.) so that it sits inside of [-1,2] or so, which is seen in step 3 above.

  3. Joe said

    So how does m(V) = m(Vi)? Since q expanded m(V) from 1 to 3.

    • James said

      Remember that V_{i} is simply V translated over some little rational distance. We haven’t “expanded” V any, we’ve simply shifted it over. It’s similar to if you had [4,5] (which has measure 1) and translated it by 2 to get [4 + 2, 5 + 2] = [6, 7], which also has measure 1. It should seem reasonable that a rigid translation of a set should not change the measure of the set.

  4. Joe said

    I think I see where I went wrong. I was thinking that V was “expanded” over all of the qi’s at once, so we’d end up with a single Vi with measure 3. So you’re saying instead that V is translated over by a single qi to get a Vi. Then we take another qi and translate over V again, and so on, so we get a countable number of Vi’s that *union* to measure 3. Is this correct?

    • James said

      Ah, no, your second sentence is correct. Yes, we will end up with lots and lots of different V_{i}‘s, each one translated over a different amount depending on what q_{i} is. And, yes, the *union* will be what we measure at the end and get the contradiction from.

  5. Joe said

    Wow. OK. It’s making sense. Step 2 seems to be the hard part. You say that x-y is rational where x in [0, 1] and y constructed from x+Q (shifted rational by some real x in [0, 1]. This seems to only make sense if it’s the same x. In other words, let x=pi. So when we say x-y is rational it would mean that pi -(pi+Q) = Q. If it were a *different* x, it doesn’t seem to work. Unless I’m completely thinking of this whole thing wrong.

    • James said

      This part is a bit easier if one knows a bit of abstract algebra. Either way, we recall how V is made. We started with a lot of sets that look like x + {\mathbb Q} for some real element x; these sets, which are the elements of the structure {\mathbb R}/{\mathbb Q} are called “cosets.” But what does this notation mean? x + {\mathbb Q} means, literally, the set of elements which are of the form x + q where x is fixed (it defines the coset here) and q\in {\mathbb Q} is any rational element. Hence, for example, in \pi + {\mathbb Q} we have elements of the form \pi + q for q\in {\mathbb Q}; in this case, we would say that \pi + \frac{1}{2} and \pi + 1 are in the same coset, since they are both in \pi + {\mathbb Q}.

      Having said this, we go back to part 2. Suppose we are given an x\in [0,1]. Then this part is saying: when we made the original Vitali set V, we had a lot of cosets w + {\mathbb Q} for w\in {\mathbb R}; x is in one of those cosets! But we may not have picked x to “represent” that coset and so x might not be in V; but it doesn’t matter, because we have picked an element y in this same coset to be in V and x is a rational distance (by construction) from y — hence, it is in one of the translates, V_{i}.

      This is tough to imagine, but pretend that we have the irrational number x =0.10110011100011110000\dots, and we have that the element y = 0.60110011100011110000\dots = x + \frac{1}{2} is in our Vitali set V. Then x = y - \frac{1}{2} and so it is an element of the coset V - \frac{1}{2} which is one of the translates V_{i} (whichever one is translated by -\frac{1}{2}. Hence, x is in the union of these V_{i}.

      • James said

        If this is too much, and is a little confusing, there are people who use many less words than I do to explain this kind of thing. Wikipedia ( has this using equivalence classes, and that might jive with your understanding. Additionally, this set is constructed in nearly every undergraduate analysis textbook [look for Vitali set or Non-measurable set] and each proof is essentially the same.

  6. Joe said

    Ah, OK. So the Vitali set V is an uncountable collection of elements between 0 and 1 that don’t necessarily contain all the real numbers between 0 and 1 (from the observation that x isn’t necessarily in V). This is where V seems to enter the twilight zone. If V did contain all of R in [0, 1], it’s measure would be 1 which wouldn’t make sense based on your step 4.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: