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Attempt to this problem: I tried in the way that firstly $A=B$ and the set is bounded then we know that the function $f(x)=m(A \cap (A+x))$ is a continuous map because $f$ is a convolution of $1_A$ and $1_{-A}$. And $f(0)>0$ so we can do it using neighborhood property.

Thinking in this way I thought that $f(x)=m(A \cap (B+x))$ is continuous, but what guarantee is there that there is an $x$ such that $m(A \cap (B+x))>0$, so we proceed like previously?

So how should I proceed? Any help is welcome. Thanks...

Glare
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Shri hari
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  • @kavi rama murthy – Shri hari Apr 14 '22 at 05:56
  • I am not him, but this answer should be helpful: https://math.stackexchange.com/questions/3819775/difference-of-lebesgue-measurable-sets

    Basically to see your function is nonzero you should integrate it, and then use the fact that an integral of a convolution $\chi_A*\chi_{-B}$ is a product of the integrals of $\chi_A$ and $\chi_{-B}$ separately.

     – Glare Apr 14 '22 at 06:58
  • The argument in the link given by Glare works: Just replace intervals by disks. – Kavi Rama Murthy Apr 14 '22 at 07:18

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