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I have problems with this exercise:

Let $(\mathbb{R}, \mathcal{B}_L,\ell) $ the Lebesgue measurement space, and assume that $Z \in \mathcal{B}_L$ such that $\ell(Z)= 0$. Prove that $\ell(Z^2)= 0$, where $$Z^2= \{x^2 : x \in Z \}$$

My attempt:

If $\{I_n\}_{n\in \mathbb N}$ is coverage of intervals of $Z$ we have that $\{I_n^2\}_{n\in \mathbb N}$ is an interval coverage of $Z^2$.

As $Z$ has zero measure then there is a coverage such that $\sum_{n\geqslant 1}\ell (I_n)<\epsilon $, for any $\epsilon \in(0,1)$

But not how to relate $\ell (I_n)$ and $\ell (I_n^2)$

Thanks

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  • $\begingroup$ I think it is straightforward to adapt this proof to your case $\endgroup$
    – Momo
    Commented Jul 19, 2021 at 19:49

2 Answers 2

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If we can prove it for the case where $Z$ is bounded, you can use the continuity of measure to prove it for the case where $Z$ is unbounded. So let us assume $Z$ is bounded, say $Z \subseteq (-M,M)$

Consider, for example, $I_n = (a_n,b_n)$ where $0\leq a_n < b_n$. Then $I_n^2 = (a_n^2,b_n^2)$ and $$l(I_n^2) = b_n^2 - a_n^2 = (b_n - a_n)(b_n+a_n) \leq (b_n-a_n)((M+1)+(M+1))$$ because any interval involved with our covering can be assumed to be contained in $(-(M+1),M+1)$. But this is a fixed constant, so summing over all the $l(I_n^2)$ we obtain $$2(M+1)\sum_n l(I_n^2) < 2(M+1)\varepsilon$$ Then let $\varepsilon \to 0$ and we get the result. There are some details you should work out based on the signs of the $a_n,b_n$, or you could split the analysis into the intersection with $[0,\infty)$ and $(-\infty,0]$.

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Start by assuming that $Z \subset (0,n)$ and let $\{I_k\}$ be a cover of $Z$ by open intervals. You can easily replace the $I_k$ by intervals $I_k' = I_k \cap (0,n)$ and still have a cover of $Z$. Express $I_k'$ as $(a_k,b_k)$.

Now, if $z \in Z$ then $z \in I_k'$ for some $k$, which by monotonicity means that $a_k^2 < z^2 < b_k^2$. It follows that the intervals $J_k' = (a_k^2,b_k^2)$ form a cover of $Z^2$.

On to the estimates. You have $$\lambda(Z^2) \le \sum_k \ell(J_k') = \sum_k (b_k^2 - a_k^2) = \sum_k (b_k - a_k)(b_k + a_k) \le n \sum_k (b_k - a_k)$$ because $(a_k,b_k) \subset (0,n)$. This means that $$\lambda(Z^2) \le n \sum_k \ell(I_k') \le n \sum_k \ell(I_k)$$ and by taking the infimum over all open covers of $Z$ you obtain $\lambda(Z^2) = 0$.

You can use countable subadditivity to generalize to the case $Z \subset (0,\infty)$ and the case of general $Z$ without too much fuss.

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