Measure Theory/Outer Measure

Outer Measure
In the previous lesson we learned that there is no total measure on all real numbers.

We also learned that, of all the properties used in the proof, we were apparently not willing to give up any of them.

But perhaps we can make progress by thinking more about that "feather" set F, used in the proof.

The problem with this set is that it is "weird", and trying to measure it causes problems.

Can we perhaps have a system of measuring intervals, and then disjoint unions of intervals, and so on -- but avoid trying to measure sets that are weird, like F?

This is precisely the strategy that we will now pursue.

We cannot currently say which sets are "weird". But we can get started on this project, by trying to measure sets in the most "real analysis way" possible: Find reasonable approximations, and then let the error go to zero.

Approximations
Recall that the most bedrock principle that we have is,

"We want open intervals to have measure equal to their length."

Therefore, it makes sense that if we have an arbitrary set $$A\subseteq \Bbb R$$, we might approximate this set by covering it with some collection of open intervals.

Whence the following definition:

Show that every subset of real numbers, $$A\subseteq\Bbb R$$, has at least one open interval over-approximation.

Hint: The collection $$\mathfrak O$$ can even be chosen to be a singleton.

True or false: The collection $$\{(-1,1),(0,2)\}$$ is an open interval over-approximation of the set $$[0,1]$$.

Estimates
For each over-approximation, there corresponds a number, an "over-estimate".

Note that because the sums here are over nonnegative terms, then the order in which the terms appear is unimportant. This is why we may merely sum over the set, rather than give it any specific indexing which would impose an ordering on the terms.

Given the open interval over-approximation $$\{(0,1),(0,2)\}$$, compute its corresponding over-estimate.

Do the same for the open interval over-approximation $$\{(0,1),(0,1/2),(0,1/4),(0,1/8),...\}$$.

Show that every over-estimate of any set is always nonnegative. (In the extended real numbers we include $$\infty$$ as a positive value.)

Outer Measure
In a sense, to measure a set, we would like to find its "least" over-estimate. This is an "approximate from above" strategy.

But a set may not have a least element, and therefore what we truly want is the infimum.

Prove that for every subset of real numbers, its outer length-measure always exists and is nonnegative. Hint: Use Exercise 1. and Exercise 4.

Initially it seems that the over-approximations may be uncountable, and that is true. This might cause us to worry that we will need to frequently compute uncountable sums, which is unwieldy.

However, there is a well-known result about uncountable sums of nonnegative quantities: Either the sum is infinity or only countably many terms are nonzero. This is shown in, for example, Terence Tao's Analysis II.

Show that for any set of real numbers, $$A\subseteq\Bbb R$$,


 * $$\lambda^*(A)=\left\{\sum_{I\in\mathfrak O}\ell(I):\mathfrak O\text{ is an countable open interval over-approximation of } A\right\}$$

Assume in your proof the following fact: Every uncountable sum is either infinity, or has at most a countable number of nonzero terms.

(The proof of this fact can be found in several places. One of them is in the second volume of Tao's analysis series.)

Because of the result above, whenever considering the outer measure of a set, we will only ever consider its countable open interval over-approximations.

Prove that $$\lambda^*$$ is translation-invariant. That is to say, for any subset $$A\subseteq\Bbb R$$ and real number $$r\in\Bbb R$$, define


 * $$A+r = \{a+r|a\in A\}$$

Now prove that $$\lambda^*(A)=\lambda^*(A+r)$$.

Hint: The basic strategy is just what you would think it is. Take any open interval over-approximation for one set and show that its translation is an open interval over-approximation for the other set. Then argue that A and $$A+r$$ have exactly the same set of over-estimates. Therefore they have the same outer measure.

Prove that $$\lambda^*$$ is monotonic. That is to say, if $$A\subseteq B\subseteq \Bbb R$$ are two subsets of real numbers, one containing the other, then


 * $$\lambda^*(A) \le \lambda^*(B)$$

Hint: Recall that if $$X\subseteq Y\subseteq\Bbb R$$ then $$\inf Y\le \inf X$$.

If you don't recall this fact, then proving it would be a good exercise to practice the fundamentals of infima.

Then show that every open interval over-estimate of B is necessarily an open interval over-estimate of A. Infer that the set of over-estimates of B, $$\mathfrak F$$ is a subset of the set of over-estimates of A, $$\mathfrak E$$. Infer that $$\inf \mathfrak E\le\inf \mathfrak F$$.

The ε/2n Trick
In this subsection, we will prove that every countable set is a null set.

The proof is a good demonstration of a tool we will need often, called the


 * "$$\varepsilon/2^n$$ trick".

Let $$A\subseteq\Bbb R$$ be a countable subset of real numbers. As such it must have an enumeration, which we may represent by the sequence $$a_1,a_2,\dots\in A$$.

Now we will build the open interval over-approximation: $$I_n = (a_n-\varepsilon/2^n,a_n+\varepsilon/2^n)$$. This is of course where the trick gets its very unimaginative name.

Notice that for each $$n\in\Bbb N$$ we have $$a_n\in I_n$$.

Also $$\ell(I_n)=\varepsilon/2^{n-1}$$.

Therefore the over-estimate which corresponds to this open interval over-approximation is


 * $$\sum_{n=1}^\infty \ell(I_n) = \sum_{n=1}^\infty \varepsilon/2^{n-1} = 2\varepsilon $$

We can now say that the set of over-estimates contains $$2\varepsilon$$ for every positive real $$\varepsilon$$. Therefore $$\lambda^*(A) \le 0$$ and since $$\lambda^*$$ is nonnegative then $$\lambda^*(A)=0$$.

This proves that A is a null set.

$$\Box$$

Confirm my claim above that


 * $$\ell(a_n-\varepsilon, a_n+\varepsilon) = 2\varepsilon $$