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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 $\varepsilon$?

Lundborg
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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 $b_i^{2} -a_i^{2} =(b_i-a_i)(b_i+a_i)$.

Kavi Rama Murthy
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  • How do you get the $\sqrt{M}$? Is it obvious that we can cover $A$ with intervals of measure less than $\varepsilon$? – Lundborg Jun 19 '20 at 12:35
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    Existence of such a cover is a basic property of Lebesgue measure. If $(a^{2},b^{2}) \subset [-M,M]$ then $b^{2}-a^{2}=(b-a)(b+a) \leq 2\sqrt M (b-a)$ because $a^{2} \leq M$ and $b^{2} \leq M$. I will let you handle the case when only part of $(a^{2},b^{2})$ intersects $[-M,M]$ . @Lundborg – Kavi Rama Murthy Jun 19 '20 at 12:43