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I am trying to do this :

Let $\lambda$ be the Lebesgue measure over $\mathbb{R}$. Let $B\in \mathcal{B}(\mathbb{R})$ such that $\lambda(B)>0$ and $A\in \mathcal{B}(\mathbb{R})$ such that $0<\lambda(A)< \infty$.

Show that $A-A$ is a neighborhood of $0$. Deduce that $B-B$ is one aswell.

I was thinking about something like this: $1_A,1_{-A}\in L^2(\mathbb{R})$. So their convolution is continuous over $\mathbb{R}$. And $1_A * 1_{-A}(0)=\lambda(A)$. We also know that the convolution is $0$ outside $A-A$ so $0\in A-A$ and since the convolution is continuous, we can find an open ball around $0$ such that its values remain in $A-A$. Is it the idea ?

I don't understand why that $B$ would be for,we have to use the $A$ we have to obtain an other set of possible infinite measure ?

Thank you.

Jose Avilez
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vadkoslo
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    I think the idea of what you're describing here is correct (even though there are some phrasing issues at the end). However what you're asking is the well known Steinhaus theorem which also has an elementary proof: https://en.wikipedia.org/wiki/Steinhaus_theorem – Just dropped in Apr 16 '22 at 18:40
  • Yes, i think i got confused at the end, i meant that we have an open ball $V$ such that for every $x\in V$, $f*g(x)>0$ this implies that there is $y$ such that $f(x-y) \neq 0$ and $g(y) \neq 0$. Then, $x-y \in A, y\in -A, x=x+y-y \in E-E$. So $V \subset A-A$. After we take a sub group $G$ of $\mathbb{R}$ of positive measure. I see that it must contains a neighborhood of $0$ (since G-G=G) but the exercise say it is in fact open in $\mathbb{R}$ i don't think i see why. I use $f=$indicator of $A $$,g=$indicator of $-A.$ Thanks for the reference i didn't know of this. – vadkoslo Apr 16 '22 at 19:04
  • I'm not quite sure after reading it. I was thinking of using a result they used, since $G$ is of finite measure, there exists an open set $U$ and $K$ compact, such that $K \subset G \subset U$ and $\lambda(U \backslash K) < \epsilon$ for all $\epsilon >0$. But i'm not sure how this forces $G$ to be an open set. – vadkoslo Apr 16 '22 at 23:31

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