# Lebesgue Measure Question

##### Active member
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

Staff member
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. ##### Active member
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. 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.**

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##### Active member
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.

##### Active member
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.