Let $A_1, A_2$ be bounded disjoint Lebesgue measurable subsets of $\mathbb R$, each of which has positive measure. Is there a translation $\tau:\mathbb R\to \mathbb R$ for which $\tau(A_1)\cap A_2$ has positive Lebesgue measure?

The question I'm actually interested in is the following: Let $A_1, A_2\subset \mathbb S^n$ be disjoint measurable subsets, each of which has positive measure. Is there a rotation $R$ of the sphere for which the measure of $R(A_1)\cap A_2$ is positive? I think I may be able to answer it if I can get some insight to the simpler question posed above.

  • $\begingroup$ Are you familiar with convolution ? $\endgroup$ – charmd Oct 22 '17 at 14:24
  • $\begingroup$ I am familiar with convolution $\endgroup$ – BindersFull Oct 22 '17 at 14:25
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    $\begingroup$ It seems the answer to the first (simpler) question is affirmative and was posted here: math.stackexchange.com/questions/870255/… $\endgroup$ – BindersFull Oct 22 '17 at 14:27
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    $\begingroup$ Essentially the same idea (but there is a more precise conclusion regarding the set of convenient translations) : see the answer here (math.stackexchange.com/questions/86209/…) $\endgroup$ – charmd Oct 22 '17 at 14:30
  • $\begingroup$ Namely, the conlusion is: there exists a non trivial open interval $I$ such that for all $x \in I$, $\tau_x(A_1) \cap A_2$ has positive measure $\endgroup$ – charmd Oct 22 '17 at 14:32

The title question was answered by this answer linked in the comments. By Fubini's theorem $\int \mu((A_1+x)\cap A_2) dx = \mu(A_1)\mu(A_2)>0,$ so some $(A_1+x)\cap A_2$ has positive measure.

To extend this to the unit sphere $S$ in $\mathbb R^{n},$ use the group of rotations $SO_{n}.$ You'll need to know how to integrate over $S$ and $SO_{n};$ there is a uniform distribution on $SO_n,$ the Haar measure, such that $g(x)$ is uniformly distributed in $S$ for each fixed $x\in S$ and for uniformly distributed $g\in SO_n$ (see for example Wikipedia). This gives $$\int_{x\in S}\int_{g\in SO_n} 1_{A_1}(g(x))1_{A_2}(x)=\mu(A_1)\mu(A_2)>0.$$ ignoring normalization, where the $1$'s denote indicator functions. Swapping the integrals gives $\mu(g^{-1}(A_1)\cap A_2)>0$ for some $g.$

As in the case of $\mathbb R$ in the comments, the conclusion can be strengthened: the map $g\mapsto \mu(g(A_1)\cap A_2)$ is continuous because $g\mapsto 1_{g(A_1)}$ is a continuous map $SO_n\to L^2(S).$ So $\{g\in SO_n\mid \mu(g(A_1)\cap A_2)>0\}$ is a non-empty open set.



You can gain some intuition in this way: If $A$ is measurable of measure $>0$, you can take a countable cover by little cubes that approximate well its measure, up to a ratio $1+ \epsilon$. This implies for every $\epsilon > 0$, there exists a small cube $C$ so that $\mu(C \cap A) > (1-\epsilon) \mu (C)$. Now, you can find a cubes of same size $C_1$, $C_2$ so that $$\mu(C_1 \cap A_1) > (1-\epsilon) \mu (C_1) \\ \mu (C_2 \cap A_2) > (1-\epsilon) \mu (C_2)$$

It should be now clear that if you translate $A_1$ so that $\tau(C_1)$, $C_2$ almost overlap, we will have an intersection of measure $>0$.


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