# Lebesgue measure. Prove that $\ell(Z^2)= 0$

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

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

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 $$(-\infty,0]$$.
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.