Measure Theory/Counter-examples

Counter-Examples
One may wonder, at this point in our investigation, a few questions which would test our understanding of length-measure.


 * Since length-measure is essentially defined by interval lengths, does that mean that if a set contains no intervals then it must have measure zero? That's certainly true for countable sets -- is it reliably true more generally?


 * If a set has an uncountable infinity of elements, does that imply that it must have non-zero measure? Effectively, this asks whether the converse of "if a set is countable then it has measure zero" is true.

In both cases, it will turn out that, shockingly, the answer is "no". And in fact, both questions can be answered by a single counter-example.

Below we will construct the ternary set, $$\mathcal C$$, and this set will have the following quite amazing collection of properties.


 * It is uncountable.
 * It is compact (closed and bounded).
 * It is measurable and has measure zero.
 * It contains no intervals.

It has many other interesting properties besides, although these are the only ones we'll have much use for right now. It will also turn out that this set becomes interesting later on when we study measurable functions, and will want similar counter-examples in that setting.

Most authors call this set the Cantor set, or some variation of that, due to its originator and one of the most profound contributors to modern mathematics: Georg Cantor. Very debatably, we may owe more to him than to any other single person in the history of mathematics, for propelling us into the modern era.

I cannot exactly pretend to know how he came upon inventing this set, except that he was probably trying to somehow construct a very weird set. His interests were quite consumed with any manor of the "weirdness" of infinity and the real numbers.

This is a great example of how "pure" interests in mathematics often come to serve many down-stream and diffuse interests throughout every other discipline of science.

The Point of a Counter-Example
A counter-example can perhaps serve a number of purposes, but here the reason why a counter-example is useful is to help you "align your intuitions". The two questions above seem to have a certain amount of intuitive appeal -- you would perhaps expect or hope that the answer is "yes".

By understanding this counter-example, you therefore correct any naive and tempting misconceptions that you might have.

In a future section, we will again look at some counter-examples. There, part of the point of those counter-examples will again be to align intuitions. However, in that setting they will also serve as a guide for building subsequent parts of the theory of measurement.

Often when mathematicians write texts, they simply show you an example, state a theorem, and give no explanation for why they are stating them. I hope that it is at least a small amount of help, for me to explain exactly why I'm including this particular information.

The Ternary Set
We will first inductively define a sequence of sets, $$\langle C_n\rangle$$, starting with $$C_0=[0,1]$$.

Next define $$C_1$$ by removing the middle third of $$C_0$$. That is to say,


 * $$C_1 = C_0\smallsetminus (1/3, 2/3) = [0,1/3]\sqcup [2/3,1]$$

Next define $$C_2$$ by removing the middle third of each interval which defines $$C_1$$. (This is stated in slightly vague language. In an exercise you will make the definition more precise.)


 * $$\begin{aligned}

C_2 &= C_1\smallsetminus (1/9,2/9) \smallsetminus (7/9,8/9) \\ &= [0,1/9]\sqcup [3/9,4/9]\sqcup [6/9,7/9]\sqcup [8/9,1] \end{aligned}$$

(Note that in the above I am "stacking" the set minus operation, meaning that we first subtract the set (1/9, 2/9) and then subtract (7/9, 8/9). Assume that when non-associative operators like this one are stacked, then you compute them "first-come-first-serve" from left-to-right.)

Proceeding likewise at each next step, we identify the intervals that compose the previous step, and remove the middle third of each.

Finally, we define the ternary set as the intersection over all sets so defined.


 * $$\mathcal C = \bigcap_{n=1}^\infty C_n$$

Prove that the two given definitions of $$\mathcal C$$ agree, by showing that $$C_1$$ as officially defined (in the green definition box) is the same as $$[0,1/3]\sqcup[2/3,1]$$.

Then let $$C_{n}$$ be as in the official definition, and consider any interval which defines it, which could be either $$I=\left[\frac{3k+0}{3^n},\frac{3k+1}{3^n}\right]$$ or $$J=\left[\frac{3k+2}{3^n},\frac{3k+3}{3^n}\right]$$ for some $$0\le k\le 3^{n-1}-1$$. Prove that by removing its middle third, one obtains, for some $$0\le k'\le 3^n-1$$, the set $$\left[\frac{3k'+0}{3^{n+1}},\frac{3k'+1}{3^{n+1}}\right]\sqcup \left[\frac{3k'+2}{3^{n+1}},\frac{3k'+3}{3^{n+1}}\right]$$.

Also show that every such $$\left[\frac{3k'+0}{3^{n+1}},\frac{3k'+1}{3^{n+1}}\right]\sqcup \left[\frac{3k'+2}{3^{n+1}},\frac{3k'+3}{3^{n+1}}\right]$$ is the result of taking some $$I=\left[\frac{3k+0}{3^n},\frac{3k+1}{3^n}\right]$$ or $$J=\left[\frac{3k+2}{3^n},\frac{3k+3}{3^n}\right]$$ for some $$0\le k\le 3^{n-1}-1$$ and removing its middle third.

Conclude that the initial definition given by removing middle thirds, is equivalent to the official definition.

Show that $$0,1,\frac 1 3\in\mathcal C$$.

Also show that $$\mathcal C$$ contains at least one element which is not the end-point of any interval which defines an nth ternary set.

Hint: at each step passing from $$ C_n$$ to $$C_{n+1}$$ one may effectively pick to "go left" or "go right".

To illustrate, going from $$C_0$$ to $$C_1$$, we can say that "going left" means choosing the interval $$[0,1/3]$$. What we mean by "choosing" this interval, is that we will find some point in it.

Next imagine "going right" after that, which would mean choosing the interval $$[2/9,3/9]$$. That is to say, while staying inside the interval chosen at the previous step, we then go left or right, to a sub-interval which is contained in the next nth ternary set.

Now consider the sequence of intervals that one obtains by making the sequence of choices summarized by writing:

L,R,L,R,L,R,L,R,...

Consider the corresponding sequence of intervals that this sequence of choices determines. Argue that the intersection of the intervals must be non-empty. There is a very famous and foundational theorem from introductory analysis which should make this a one- or two-line proof.

Then you only need to argue that any element in this intersection is not an end-point of any interval which defines an nth ternary set.

Show that the ternary set contains no intervals. Hint: Suppose for contradiction that it does contain an interval, I. Then show that, eventually, some nth ternary set's defining intervals must be shorter in length than I.

Prove that the ternary set is uncountable, by demonstrating an injective map from the set of real numbers [0,1] to $$\mathcal C$$.

Hint: Every number in [0,1] has a binary representation, which is just a sequence of 0s and 1s. Bijectively map each sequence to a sequence of "left" and "right" choices, as in ''Exericse 2. Ternary Warmup'', which then maps injectively to a point in $$\mathcal C$$.

(We only really need to show that this map is injective in order to decide that $$\mathcal C$$ is uncountable. Because $$\mathcal C$$ contains no intervals, though, then any sequence of left and right choices actually must uniquely correspond to a number in $$\mathcal C$$.  This actually makes the map "injective in both directions", so to speak.  Which is to say, the map is a bijection.  Feel free to prove this, or don't, as you prefer.)

It is clear that $$\mathcal C$$ is bounded. Show further that it is closed. The proof should be extremely short, maybe just one sentence.

Show that $$\lambda(C_n) = (2/3)^n$$.

Then use the continuity of measure to prove that $$\lambda(\mathcal C)=0$$.