Measure Theory/Outer Measure

Outer Measure

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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

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Recall that the most bedrock principle that we have is,

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

Whence the following definition:

Definition: over-approximation

Let   and let   be a collection of open intervals such that  . Then we call   an open interval over-approximation of A .

Exercise 1. Every Set Has One

Show that every subset of real numbers,  , has at least one open interval over-approximation.

Hint: The collection   can even be chosen to be a singleton.

Exercise 2. Can They Overlap?

True or false: The collection   is an open interval over-approximation of the set  .

Estimates

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

Definition: over-estimate

Let   be any open interval over-approximation of any set. Then we define   to be the over-estimate corresponding to  .

If any term is   or if the sum diverges, then we say that  . In general,  .

Because there is no requirement that   be countable, we therefore also need to define the sum over an uncountable set of nonnegative numbers.

Let   be any set of nonnegative real numbers. We define the sum over X by the supremum taken over all countable sums.

That is to say, let

 

Then the sum over X is

 

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.

Exercise 3. Over-estimate Something

Given the open interval over-approximation  , compute its corresponding over-estimate.

Do the same for the open interval over-approximation  .

Exercise 4. Estimates Bounded Below

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

Outer Measure

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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.

Definition: outer measure

Let   be any set of real numbers. Let   be the set of all over-estimates of A. We then define the outer length-measure of A to be

 

As we will prove below, the outer measure is a set function taking values in  .

 
Exercise 5. Always There

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.

Exercise 6. Outer Measure Is Determined by Countable Sums

Show that for any set of real numbers,  ,

 

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.

Exercise 7. Translation Invariant

Prove that   is translation-invariant. That is to say, for any subset   and real number  , define

 

Now prove that  .

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   have exactly the same set of over-estimates. Therefore they have the same outer measure.

Exercise 8. Monotonicity

Prove that   is monotonic. That is to say, if   are two subsets of real numbers, one containing the other, then

 

Hint: Recall that if   then  .

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,   is a subset of the set of over-estimates of A,  . Infer that  .

Definition: null set

Let   be any subset of real numbers. If   then we say that A is a null set.

The ε/2n Trick

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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

"  trick".

Let   be a countable subset of real numbers. As such it must have an enumeration, which we may represent by the sequence  .

Now we will build the open interval over-approximation:  . This is of course where the trick gets its very unimaginative name.

Notice that for each   we have  .

Also  .

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

 

We can now say that the set of over-estimates contains   for every positive real  . Therefore   and since   is nonnegative then  .

This proves that A is a null set.

 


Exercise 10. Confirm the Length Please

Confirm my claim above that