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Lebesgue Measure Question

TheBigBadBen

Active member
May 12, 2013
84
I have a final coming up, so I thought I'd post some of my review questions as a way of checking my work. I think I have a working answer for this one, but I'm not sure it's totally right. I'll post it upon request.

At any rate, two related questions:

(1)
Suppose that \(\displaystyle E \subset \mathbb{R}\) is a set such that \(\displaystyle m^*(E)=0\). Prove that \(\displaystyle m^*(E^2)=0\), where \(\displaystyle E^2 = \{x^2|x\in E\}\)

(2)
Suppose that \(\displaystyle f:\mathbb{R}\rightarrow\mathbb{R}\) is a K-Lipschitz function. Show that \(\displaystyle m^*(E^2)≤Km^*(E)\) for all \(\displaystyle E\subset\mathbb{R}\)

Note that \(\displaystyle m^*\) refers to the Lebesgue outer-measure.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I think it would be better if you post your work, and then our helpers can give you a critique.

You are more likely to get help this way rather than to request proofs be given to which you can compare your work. :D
 

TheBigBadBen

Active member
May 12, 2013
84
I think it would be better if you post your work, and then our helpers can give you a critique.

You are more likely to get help this way rather than to request proofs be given to which you can compare your work. :D
Fair enough. My proof for the first:

We define the outer measure by

\(\displaystyle m^*(E)=inf \left\{ \sum_{n=1}^{∞}\ell(I_n):E\subset \cup_{n=1}^{∞} I_n \right\}\)

Where each \(\displaystyle I_n\) is an open interval. Since the outer measure of E is 0, we can state that for any \(\displaystyle \epsilon >0\), we can equivalently find a collection of sets {I_n} so that the sum of their lengths is less than \(\displaystyle \epsilon\) and such that E is contained in their union.

We begin with the case that \(\displaystyle E\subset [0,\infty)\)

Consider any \(\displaystyle \epsilon>0\). There exists a collection of sets {I_n} so that the sum of their lengths is less than \(\displaystyle \sqrt{\epsilon}\) and such that E is contained in their union. We note that for any \(\displaystyle x\in I_n\), we have \(\displaystyle x^2 \in I_n^2\). It follows that \(\displaystyle E^2\subset \cup_{n=1}^{∞} I_n^2\).

Now, for \(\displaystyle I_n=(a_n,b_n)\), we note that \(\displaystyle I_n^2=(a_n^2,b_n^2)\), from which it follows that
\(\displaystyle \ell(I_n^2)=b^2-a^2 = (b - a)(b + a) \geq (b - a)^2 = \ell(I_n)^2\)
^^^^
Not as useful as I thought...


It follows that


\(\displaystyle \sum_{n=1}^{\infty}\ell(I_n^2)≤\)
\(\displaystyle \sum_{n=1}^{\infty}\ell(I_n)^2≤\) <--- this is wrong
\(\displaystyle \left[ \sum_{n=1}^{\infty}\ell(I_n) \right]^2<\)
\(\displaystyle \left[ \sqrt{\epsilon} \right]^2=\epsilon\)


We conclude that for \(\displaystyle E\subset [0,\infty)\), \(\displaystyle m^*(E^2)=0\).

Takes a while to type up...
Any critique so-far? Nit-picking is welcome here!
I think I should be able to generalize this relatively easily to the entire real line, but if anybody has a cleverer method I readily welcome that.

**So I found a flaw in my own proof, where I try to establish the inequality. Ends up, I have it going the wrong way. I need to find a way to fix that.**
 
Last edited:

TheBigBadBen

Active member
May 12, 2013
84
My idea for the proof for the other one is similar. The idea is that if f is Lipschitz, then for any interval I, we should have
\(\displaystyle m^*(f(I))≤K\ell(I)\)
Still not sure how to put it all together.
 

TheBigBadBen

Active member
May 12, 2013
84
Figured out the solution to my problem. I had tried to come up with some sort of inequality, and wrote:

We begin with the case that \(\displaystyle E\subset [0,\infty)\)...

Now, for \(\displaystyle I_n=(a_n,b_n)\), we note that \(\displaystyle I_n^2=(a_n^2,b_n^2)\), from which it follows that
\(\displaystyle \ell(I_n^2)=b^2-a^2 = (b - a)(b + a) \geq (b - a)^2 = \ell(I_n)^2\)
Which, as you can see later, is the wrong sort of inequality for what I wanted to do. As it ends up, the easier thing to do is to start by looking at bounded intervals, eventually noting that the arbitrary union of null sets is a null set. So, I'd have something like:

Define \(\displaystyle E_N = E \cap [0,N], N\in \mathbb{N}\). We note that
\(\displaystyle I_n^2=(a^2,b^2)\), and
\(\displaystyle \ell(I_n^2)=b^2-a^2 = (b - a)(b + a) \leq 2N(b-a) = 2N\ell(I_n)\)

From there, it's easy enough to use cascading inequalities to show that an interval covering \(\displaystyle \{I_n\}_{n\geq 1}^2\) of \(\displaystyle E_N^2\) can be made arbitrarily small. Following a similar logic, we can do this for an integer \(\displaystyle -N\), defining

\(\displaystyle E_{-N} = E \cap [-N,0], N\in \mathbb{N}\)

And producing the same result. Since \(\displaystyle E^2\) is the union of all sets \(\displaystyle E_N^2\), we deduce that \(\displaystyle E^2\) is the countable union of null sets, and hence is itself a null set.

I'm pretty happy with this proof, but if someone can offer something better, I'm listening.