This is a follow up to this question: Upper bound "square" of Lebesgue measure of set
Let $\lambda$ be the Lebesgue measure, $A$ a set with $\lambda(A) < \varepsilon$ and $M > 0$. Consider the set $A^2 = \{a \cdot a \ | \ a \in A\}$. Can we produce an upper bound on $\lambda(A^2 \cap [-M, M])$ in terms of $\varepsilon$?
Yes, $\lambda (A^{2}\cap [-M,M]) \leq 2\sqrt M \epsilon$. To see this note that we can cover $A$ by intervals $(a_i,b_i)$ with $\sum (b_i-a_i) <\epsilon$. Now note that $A^{2}\cap [-M,M]$ is convered by $(a_i^{2},b_i^{2})\cap [-M,M]$ and $\lambda (a_i^{2},b_i^{2})\cap [-M,M]) \leq 2\sqrt M (b_i-a_i)$ since $b_i^{2} -a_i^{2} =(b_i-a_i)(b_i+a_i)$.