Measure Theory/Monotone Functions Differentiable

Monotone Functions

edit

Differentiable A.E.

edit

Although a monotonically increasing function might fail to be differentiable at uncountably many points, the next-best thing would be for the set of "failure points" to have measure zero. That is to say, we might instead hope that a monotonically increasing function is differentiable almost everywhere.

Indeed this turns out to be true! However, it will require a significant journey to prove that it is true.


Exercise 1. Not Just Monotone


Recall that every function of bounded variation can be decomposed into a difference of two monotonically increasing functions. Now assume that every monotone function is differentiable a.e. Prove that every function of bounded variation is differentiable a.e.

Assume throughout this lesson that   is monotonically increasing on the compact interval [a,b].

Where the Upper Derivative Is Infinite

edit

One way in which the derivative may fail to exist at x is for  . We will prove first that the set of all points at which this happens is null.

To approximate this set, we first define the set  , which is effectively the set of points at which the upper-right derivative is "large".

We would like to show that   becomes small as c becomes large.

In order to do so, we can recall the mean value theorem, which tells us that   for some x in the interval, if f is differentiable on (a,b). If   is a lower bound on this derivative, then  .

The left-hand side, b-a, is the measure of the set [a,b], which in our setting is like  . Inspired by this, we will try to prove

 

Let   and consider the family of all intervals where the inequality holds.

 

The reason for considering such a collection is that it may be easier to measure, as a collection of intervals, than the more "chaotic" set  . Of course to be "effective" we will need this collection to cover  .

Unfortunately the set   may not cover  , and if you set out to prove that it does, you will notice a difficulty in one step.

However, this difficulty is resolved if, instead of   we let   and then define the collection

 

Then   is a cover, as you will demonstrate in the exercise below.

Notice that if   then by definition   which puts some "space" between c' and  . Therefore there is some   such that  , therefore there is some   for which  .


Exercise 1. The Need for Space


Use the observations above to prove that, for each   and for each   there is some   such that   and  . Hint: Consider the interval  .

However, there is even a problem with  . This collection will have a ton of pathological overlap between the various intervals which it contains.

Therefore, what we really seek is to select from it some finite, disjoint sub-collection which is close enough to  .

Vitali Coverings

edit

The kinds of concern above occurs often enough that it is worth handling generally.

Definition: cover in the sense of Vitali

Let   be any subset, and   a collection of nondegenerate compact intervals. We say that   covers E in the sense of Vitali if the following two condition is met.

For any   then for every   there exists an interval   such that

1)  

2)  .


Exercise 2. Prove Vitali's Lemma


Assume E has finite outer-measure and   covers E in the sense of Vitali. Show that for any   there exists a finite disjoint sub-collection   such that

 

Here are some guiding steps.

1) Let   be an open set such that   and  .  (If it is not clear, then you may want to prove that such a set exists.)
2) Construct the collection   where   if  .  Argue that   also covers E in the sense of Vitali.
3) Inductively construct a finite collection of   by first picking an arbitrary  .  
4) Next construct, for each   and collection  , the collection of all intervals in   which are disjoint from all of the intervals so far.
  
Also construct the set of all lengths of these intervals in   and argue that this is a nonempty set of real numbers bounded above, and therefore has a supremum, call it  .  Infer that there is some   with  .
5) For each   argue that  .
6) The path home should be not very hard to find from here.  


Exercise 3. Explain O

Explain why the open set   was necessary in the proof above.


Exercise 4. Actually There's More to Vitali's Lemma

We have proved Vitali's Lemma for the purposes of proving our theorem about monotone functions. However, there is a further property that this finite collection,  , has which may prove useful later. We might as well observe and prove it now, while we're here.

Prove that  . Recall that the notation   means the interval with the same center as   but five times the width.

Suggestions for how to proceed: Let   and of course if   then the proof is finished.

1) Observe that if   then the proof is finished.  Otherwise we only need to consider the case that  .
2) Infer that there is an   with  .
3) Note that although we only took the intervals   in the exercise above, the sequence of intervals continues beyond this.  Argue that there must be some   which intersects I.  Hint: If this were false there would be a positive lower bound on the sequence  .  Call N the least index for which   intersects I.  
4) Argue that   and infer that  . 

Finishing the Proof

edit

Returning to the proof from before, use the fact that   is a Vitali covering to construct a finite disjoint collection of intervals,

 

such that  .


Exercise 5. Finish the Case for Infinity


1. Decompose   into two parts, the part inside   and the part outside. Use subadditivity to infer that

 

Then let   and  .

2. Now use the above, together with the continuity of measure, to conclude that the set of points at which the upper-left derivative is infinite.

Then argue that the same is true for the lower-right, upper-left, and upper-right derivatives.

Unequal Derivatives Almost Nowhere

edit

We now go after the other way in which a function might fail to be differentiable, which is for the derivatives to be finite but unequal. Again we will focus initially on the right-handed side but will later generalize to the inequality of any of the derivatives.

That is to say, we will show that   has measure zero. Very similar to the previous proof, we would like to build a countable sequence of sets which approximate this set and then use the continuity of measure at the end.

Just as before, we need a bit of space between the two derivatives (although this time we will look at what happens as the space becomes arbitrarily close to zero). Therefore we will define

 

Let   and we will try to show that   is small.

As in the previous part regarding where the derivatives go to infinity, we will leverage the Vitali Covering Lemma. It will help later if, moreover, this is situated inside of some open set.


Exercise 6. Find O


Show that there is an open set   such that   with the property that  .

Then define a collection of closed intervals that are each inside of  , in a way analogous to what was done for the infinite derivative case. Argue that this is a cover in the sense of Vitali.

Complete rest of the argument in a way that is strongly parallel to the proof for the case of infinity, until you reach

 

and then justify and use the fact that  .

The path to the final result should not be very hard (nor very easy!) to find from here. But do your best.