## Mild Introduction.

Last time, we described some facts about Lebesgue measure and showed that despite being a nice measure of sets on the real line there are some sets which cannot be measured by the Lebesgue measure.  Intuitively, this gives us a big contrast with “real world” objects — there are very few things (except for the very, very small and the very, very large, perhaps) which we cannot hold a ruler next to and measure or, at least, approximate.  The previous set we constructed could not even be approximated — or, at least, it is not obvious how to approximate such a beast!

On the other hand, we will show that there are sets which, informally speaking, the Lebesgue measure “doesn’t care” about.  These sets are called sets of measure zero, and they’re quite easy to describe.

Definition.  We say that $A$ is a set of measure zero if $\mu(A) = 0$.

These sets will act like a “zero” element with measures.  If we have some set $B$, then it will turn out that if $A$ has measure zero, we will show that $\mu(B) = \mu(B\cup A)$.  This is more or less what I mean when I say that the Lebesgue measure does not care about these sets.

[Note: from now on, I will use the term “measurable” to mean “Lebesgue measurable” and I will qualify when I don’t mean this.]

## Sets of Measure Zero and Why They Are Kind of Cool.

One thing that we saw last time is that sometimes sets aren’t measurable.  In particular, it is usually not trivial to show that some arbitrary set is measurable if it is not an elementary set.  You might recall the “outer measure only” definition of measurability that we defined last time (recall, we are only working in the unit interval for now):

A set $A$ is measurable if and only if $\mu^{\ast}(A) + \mu^{\ast}([0,1] - A) = 1$.

Using this, we can prove something pretty neat about sets of measure zero; namely, we will prove that if their outer measure is zero, then they are actually Lebesgue measurable with Lebesgue measure zero!  Should this really be surprising?  If outer measure is a kind of measure of “thickness”, then having no thickness at all should give us a similar Lebesgue measure.  But this interpretation is not exact and, in fact, I still consider this theorem as one of the more “amazing” theorems in measure theory, not only due to the interesting mental image it presents but also due to its usefulness in future proofs.    Let’s do it.

Theorem.  If $A$ is a subset of the unit interval and $\mu^{\ast}(A) = 0$, then $A$ is measurable with $\mu(A) = 0$.

What’s nice about this kind of theorem is that we don’t have all that much to work with currently, so it is a great proof to give students just beginning measure theory.  I feel a bit bad demonstrating the proof on this blog, but because it is readily available in many places (including almost every book on measure theory I have) I will provide it here.

Proof.  Note that $[0,1] \subseteq (A \cup ([0,1]-A)$ and so we have

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

but $[0,1] - A\subseteq [0,1]$, and so, in fact

$\mu^{\ast}([0,1] - A) \leq 1$

which, in total, implies that $\mu^{\ast}([0,1]-A) = 1$, proving the theorem.  $\Box$

What was the important part here?  It was the fact that we could use monotonicity the second time.  Since our equation only depended on $[0,1] - A$‘s measure, we could use the fact that this was a subset of $[0,1]$.  Like I said: it’s almost like $A$ doesn’t even matter in this case; it just vanishes under the measure!  Here’s a nice related result:

Theorem.  If $B$ is any set and $A$ is a set of measure zero, then $\mu^{\ast}(B) = \mu^{\ast}(B\cup A)$.

Now, this theorem is the formal statement that if $A$ has measure zero, then the Lebesgue measure doesn’t care about it.

Proof.  We’ll do the same sort of trick as the last theorem.  Note that $B\subseteq B\cup A$ so we have $\mu^{\ast}(B) \leq \mu^{\ast}(B\cup A)$.  But note that subadditivity gives us $\mu^{\ast}(B\cup A)\leq \mu^{\ast}(B) + \mu^{\ast}(A)$.  This gives us the following inequality in total:

$\mu^{\ast}(B) \leq \mu^{\ast}(B\cup A)\leq \mu^{\ast}(B) + \mu^{\ast}(A) = \mu^{\ast}(B)$

from which we conclude that $\mu^{\ast}(B) = \mu^{\ast}(B\cup A)$$\Box$

This theorem is rather nice.  But the critical reader will point out something: we have not actually shown that such sets of measure zero exist!  We may be talking about some object which does not actually occur.  We ought to give some examples now, lest the reader think that such things never come up in real life.

[This type of idea is often the subject of math horror stories.  One such one goes like this: some bright grad student was defending his thesis and had proven a whole number of interesting things about a certain space.  At the end of his talk, he was asked for an example of such a space, and the grad student could not think of one.  One distinguished professor who is at the defense speaks up and notes that the only space satisfying such properties is the trivial space.  The grad student has therefore said much about nothing.]

## Examples of Sets of Measure Zero.

First, we’ll start with some easy examples, and then we’ll show an interesting result.

A single point has measure zero.  This is perhaps not so surprising.  This one is easy to see: if $p$ is the point, then $(p -\epsilon, p+\epsilon)$ for sufficiently small $\epsilon > 0$ will cover $\{p\}$ and has length which goes to zero.

A finite number of points has measure zero.  Same sort of idea.  Let’s let $p_{1}, \dots, p_{n}$ be the points.  Then $\bigcup_{i=1}^{n} (p_{i} - \epsilon, p_{i} + \epsilon)$ covers this set for sufficiently small $\epsilon >0$ and the total measure of this is less than or equal to $n\epsilon$ which, since $n$ is finite, goes to 0.

At this point, the excited reader might want to go on to say a countable number of points is of measure zero.  And it is true.  But the proof requires a little trick that we’ve used before, so we denote this as a proposition.

Proposition.  A countable number of points has measure zero.

Proof.  Let’s let these points be $\{p_{i}\}_{i=1}^{\infty}$.  Then we want to do something like the finite number of points argument, but we need to be clever because adding up infinitely many small things might give us something which is really big.  So we do the following.  Set $\epsilon > 0$, and now cover each point with the open interval $\displaystyle (p_{i} - \frac{\epsilon}{2^{i+1}},p_{i} +\frac{\epsilon}{2^{i+1}})$.  Note now that when we union these together, we get

$\mu^{\ast}\left(\bigcup_{i}(p_{i} - \frac{\epsilon}{2^{i+1}},p_{i} +\frac{\epsilon}{2^{i+1}})\right) \leq \sum_{i=1}^{\infty} \frac{2\epsilon}{2^{i+1}} = \sum_{i=1}^{\infty}\frac{\epsilon}{2^{i}} = \epsilon$

and since $\epsilon > 0$ was arbitrary, we get that $\{p_{i}\}_{i=1}^{\infty}$ has measure zero.  Nice.  $\Box$

Corollary.  The rational numbers in the unit interval has measure zero.

Is it true that an uncountable number of points has measure zero?  Not necessarily; the interval $[0,1]$ is made up of an uncountable number of points, but it has measure 1.  Are there sets with an uncountable number of points that have measure zero?  Yes.  For example, the Cantor Set has measure zero, but it has an uncountable number of points.  A quick argument (which has to be filled in to make rigorous) is something like this: every time we remove an interval, it’s of size $\frac{1}{3^{n}}$ and there are $2^{n-1}$ of them (the first time we remove 1 of size $\frac{1}{3}$, the second we remove 2 of size $\frac{1}{9}$, and so on…).  Thus, we begin with an interval of length 1, and the final length (which is the measure of the Cantor set itself) will be

$\displaystyle 1 - \sum_{i=1}^{\infty}\frac{2^{i-1}}{3^{i}}$

but this is a geometric series!  So, the solution is

$\displaystyle \frac{\frac{1}{3}}{1 - \frac{2}{3}} = 1$

which gives us that the length of the Cantor set is actually zero.  Even if we didn’t know it had measure zero, the Cantor set is the intersection of the complement of countably many closed intervals, so it would have to be measurable — but now we have verification not only that it is measurable, but also that its measure is zero!

Proposition.  The irrational numbers in the unit interval are measurable and they have measure 1.

There are a few cute tricks to show that the irrational numbers are measurable, but my favorite way is to show a slightly more general property and go from there.  Let’s just do that real quick.

Lemma.  If $A$ is measurable in the unit interval, then so is $[0,1] - A$.

Proof.  If $[0,1]-A$ were to be measurable, then we would need $1 = \mu^{\ast}([0,1] - A) + \mu^{\ast}([0,1]-([0,1]-A))$.  But note that $[0,1] - ([0,1] - A) = A$ and so this reduces to $1 = \mu^{\ast}([0,1] - A) + \mu^{\ast}(A)$ which we know to be true since $A$ is measurable.  $\Box$

Proof of the Proposition.  So how can we apply this?  Well, we know the rationals have measure zero, so the complement (the irrationals) is also measurable!  Good.  Let $A$ be the rationals in the unit interval.  Then $[0,1]\subseteq A\cup ([0,1]-A)$ and also $[0,1] - A \subseteq [0,1]$, so we have

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

which shows the measure of the irrational numbers in the unit interval is 1.  $\Box$

## Next Time.

Next time, we’ll go over some function stuff.  This measure business is nice and all, but we ultimately want to talk about integration, so it would be nice to have some way to talk about functions that bring measurable sets to measurable sets.  In many ways, this will be a similar deal to how continuous functions force open sets in the range to have open pre-images.

### 9 Responses to “Real Analysis Primer, part 4: Sets of Measure Zero and Some Measure Problems.”

1. Angie said

I know it has been a while since this blog entry was made, but I found it very helpful. I was particularly interested in the corollary that the rational numbers have measure zero. I am working on a homework problem for my analysis and this seems helpful. I am trying to show that the set of rationals can be squeezed into a set of intervals whose total length is less than epsilon, where epsilon is any fixed positive number.

2. Anonymous said

Countable sets have outer measure Zero. But if a set A is of measure zero, can we say it is countable? Any example or counter-example, please.

• James said

This sounds like one of the homework questions I had to do in analysis, so I’ll give a hint without explicitly saying it. Countable sets in, say, ${\mathbb R}$ do indeed have (Lebesgue) measure zero. For an uncountable set, you need to work a little harder — think about a set which is constructed by starting with an interval and “taking out” a lot. There’s a pretty famous example of this.

• Anonymous said

no, we cant say that a set A with measure zero is countable. cantor set is an example of a uncountable set with measure zero.

• muneera m i said

A set A with measure zero need not be countable, a wonderful example is contor set which has measure zero , but the set is uncountable.

• James said

Yes, I didn’t want to spoil it, but the cantor set is an example of a set which is uncountable but has measure zero. If you’re brave, think of some other ones!

3. owino Joseph Ochieng said

would please assist me: 1: prove that the Cantor-ternary set is measurable and calculate its measure?
2: Explain why the Cantor-ternary set is not countable?
3: every countable set has measure zero. prove that every set of measure zero is not necessarily countable.