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If $E$ has measure zero, then does $E^2$ have measure zero?

I'm trying to find a proof for the following question: Suppose $A\subset\mathbb{R}$ is a set of measure zero. Show that the set $A^2=\{x^2 \in \mathbb{R} \,|\, x \in A\}$ is also a set of measure zero.

I feel like this is a very easy proof and I am just missing something. I don't think stating "the map $f:A\to A^2$ also maps the collection of intervals covering $A$ to a collection of intervals covering $A^2$" is complete. I don't believe this qualifies as a Lipschitz function, since the derivative of the function $f$ is unbounded on $\mathbb{R}$. Please help, is there something I am missing here?

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marked as duplicate by Qiaochu Yuan Sep 7 '12 at 5:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

up vote 1 down vote accepted

What we are looking for is whether $m(f(A)) = 0$ where $f(x) = x^2$. For simplicity we can think of $A$ as just consisting of positive members. Define $A_n = A \cap [0, N]$. All we need to show is that $m(f(A_N)) = 0$ since $f(A) = \bigcup_n(f(A_n))$, and a countable union of measure zero sets is measure zero.

Now, $A_N \subset A$ so it is a set of measure zero (I am assuming Lebesgue measure). Therefore, for all $\epsilon > 0$, there exists $\{I_n\}$ (a collection of open intervals) such that $A_N \subset \bigcup_n{I_n}$ and $\sum_n{m(I_n}) < \frac{\epsilon}{2N}$. Let $I_n = (a_n,b_n)$ so $m(I_n) = b_n-a_n$. Thus, $$m(f(I_n)) = b_n^2 - a_n^2 = (b_n-a_n)(b_n+a_n).$$ We see that $$\sum_n{m(f(I_n))} = \sum_n{(b_n - a_n)(b_n + a_n)} \leq \sum_n{(b_n - a_n)*2N} = \sum_n{m(I_n)*2N} < \epsilon$$ Thus, $m(f(A_n)) = 0$ which means $f(A) = 0$.

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It is not globally Lipschitz, but it is locally Lipschitz. Since $m(X) = \lim_{R \to \infty} m(X \cap [-R,R])$ this is enough.

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