TITLE: Maps from sets of measure zero to sets of measure zero
QUESTION [7 upvotes]: Possible Duplicate:
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?

REPLY [3 votes]: 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 $f(A) = 0$.