MHB Proving Null Sets: Lebesgue Measure and Lipschitz Functions

[TheBigBadBen]

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

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

Note that $$m^*$$ refers to the Lebesgue outer-measure.

 

[MarkFL]

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]

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

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

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

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

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

Now, for $$I_n=(a_n,b_n)$$, we note that $$I_n^2=(a_n^2,b_n^2)$$, from which it follows that
$$\ell(I_n^2)=b^2-a^2 = (b - a)(b + a) \geq (b - a)^2 = \ell(I_n)^2$$
^^^^
Not as useful as I thought...It follows that$$\sum_{n=1}^{\infty}\ell(I_n^2)≤$$
$$\sum_{n=1}^{\infty}\ell(I_n)^2≤$$ <--- this is wrong
$$ \left[ \sum_{n=1}^{\infty}\ell(I_n) \right]^2<$$
$$\left[ \sqrt{\epsilon} \right]^2=\epsilon$$We conclude that for $$E\subset [0,\infty)$$, $$m^*(E^2)=0$$.

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.**

 

[TheBigBadBen]

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
$$m^*(f(I))≤K\ell(I)$$
Still not sure how to put it all together.

 

[TheBigBadBen]

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 $$E\subset [0,\infty)$$...

Now, for $$I_n=(a_n,b_n)$$, we note that $$I_n^2=(a_n^2,b_n^2)$$, from which it follows that
$$\ell(I_n^2)=b^2-a^2 = (b - a)(b + a) \geq (b - a)^2 = \ell(I_n)^2$$

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

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

$$E_{-N} = E \cap [-N,0], N\in \mathbb{N}$$

And producing the same result. Since $$E^2$$ is the union of all sets $$E_N^2$$, we deduce that $$E^2$$ 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.