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$G$ is a locally compact Hausdorff topological group, $m$ is a (left) Haar measure on $X$, $A$ and $B$ are two finite positive measure in $G$, that is $m(A)>0$, $m(B)>0$.

My question is:

Can we conclude that $AB= \{ab, a\in A, b\in B\}$ contains some non-empty open set of G?

Is this question right? Or is this right just for $G=R^n$, $R^n$ is the Euclid space, and $m$ is the Lebesgue measure on $R^n$. If so, how to prove it?

Thanks a lot.

Alex Ravsky
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David Chan
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  • Do you mean that $A$ and $B$ are subgroups or something? Otherwise, in $R$, why can't you have $A = B = [1,2]$, with $A+B = [2,4]$ not a neighborhood of identity? –  Apr 30 '15 at 04:15
  • @ Mike Haskel: Thanks, I have modified it. – David Chan Apr 30 '15 at 04:18
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    Looking at proofs of the theorem for $\mathbb{R}^n$ (e.g. here), they appear, at least on first glance, to use reasoning about metric spaces in a non-trivial way. It may be possible to cut out this assumption by translating those parts of the argument into language about uniform covers, or the result may only hold of Polish groups (or similar). I'll have to think about it more (or someone else will beat me to it :-)). –  Apr 30 '15 at 04:51
  • This result is sometimes called Steinhaus's theorem. https://en.wikipedia.org/wiki/Steinhaus_theorem – Sean Eberhard May 03 '15 at 10:50

3 Answers3

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Here is another proof, using regularity of the measure instead of convolution.

Claim: The result holds when $B=A^{-1}$.

Proof: By regularity there is a compact set $K$ and an open set $U$ such that $K\subset A\subset U$ and such that $m(U)<2m(K)$. The multiplication map sends $\{1\}\times K$ into $U$, so by continuity of multiplication and compactness of $K$ there is a neighbourhood $V$ of $1$ such that multiplication sends $V\times K$ into $U$. But then if $x\in V$ the sets $K$ and $xK$ are each more than half of $U$, so $K\cap xK$ is nonempty, so $x\in KK^{-1}$. Thus $KK^{-1}$ contains a neighbourhood $V$ of $1$. $\square$

Claim: The result holds in general.

Proof: By regularity we may assume both $A$ and $B$ are compact. For $x$ running over $G$ we have $$\int m(A\cap xB^{-1}) \,dx = \int\int 1_A(y) 1_B(y^{-1}x) \,dy\,dx = m(A)m(B)>0$$ by Fubini's theorem, so there is some $x$ such that $m(A\cap xB^{-1})>0$. Now apply the previous result to $A\cap xB^{-1}$. Since $$(A\cap xB^{-1})(A\cap xB^{-1})^{-1} \subset ABx^{-1},$$ we deduce that $AB$ contains a neighbourhood of $x$. $\square$

Sean Eberhard
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  • Wow, You are really too much! – David Chan May 12 '15 at 01:58
  • Why can we use Fubini's theorem? G is not a sigma-finite measure space... – David Chan May 12 '15 at 08:11
  • The integrand is zero for $(x,y)$ outside the compact set $AB\times A$, so such problems do not arise. Alternatively you can always just restrict to a subgroup $H$ of $G$ generated by a compact neighbourhood of the identity: this is an open $\sigma$-compact subgroup. – Sean Eberhard May 12 '15 at 08:48
  • :What dose this H mean, Then? – David Chan May 12 '15 at 12:46
  • I just mean if you take any compact neighbourhood $K$ of the identity and let $H$ be the subgroup of $G$ generated by $K$ then $H$ is an open $\sigma$-compact subgroup of $G$. The Haar measure of $G$ is then just the sum of its restrictions to the cosets of $H$, so by translating $A$ and $B$ we may assume they intersect $H$ in sets of positive measure. In effect we may assume that $G$ is $\sigma$-compact so certainly $\sigma$-finite. – Sean Eberhard May 12 '15 at 12:54
  • This is a slick proof. As a matter of fact, I was wondering if there is a nice solution without (any) integrals. But until now, I can't find it...Oh my god... – David Chan May 15 '15 at 08:17
  • About the regularity: If we define the Haar measure as a Baire measure, there's of course no problem. But if we define the Haar measure as a Borel measure (translation invariant, finite on compact subsets, nonzero on nonempty open sets, outer regular), then it need not be inner regular. If $A$ isn't $\sigma$-finite, it can be that $\mu(K) = 0$ for all compact $K\subset A$, but $\mu(A) = +\infty$. – Daniel Fischer Dec 13 '16 at 12:20
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You can prove this by considering the convolution $f(x) = \int_G \chi_{A}(xy^{-1}) \chi_{B}(y) dy$ of the characteristic functions of $A, B$. WLOG, we can take $A, B$ to be compact sets of positive measure. So $f$ is continuous and has positive integral. That $AB$ has interior immediately follows.

Alternatively, you can also use the regularity of Haar measure to show this directly.

Sean Eberhard
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convoluted
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  • "That AB has interior immediately follows." I don't understand this. Would you elaborate on it? – Makoto Kato May 03 '15 at 00:56
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    @WilliamCurtis Choose some $x_0$ such that $f(x_0)>0$. There is some neighbourhood $U$ of $x_0$ such that $f>0$ on $U$. That means that for every $x\in U$ there is some $y$ such that $xy^{-1}\in A$ and $y\in B$; in other words $x\in AB$. – Sean Eberhard May 03 '15 at 10:49
  • Alternatively, you can also use the regularity of Haar measure to show this directly."What does this mean? Could you give a proof avoid using convolution way?As matter of fact, I always would like an original proof. Also, this answer is perfect, too. – David Chan May 04 '15 at 23:46
  • @Sean Eberhard: Why f is continuous? – David Chan May 11 '15 at 10:01
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    @DavidChan Approximate $\chi_A$ and $\chi_B$ in $L^1$ norm by continuous and compactly supported functions. After convolution the $L^1$ error becomes an $L^\infty$ error, so $\chi_A\ast\chi_B$ is a uniform limit of continuous functions, so continuous. The convolution operation has many fundamental smoothing properties like this one. – Sean Eberhard May 11 '15 at 12:00
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You can take this example if the topological group $G$ has finite Haar measure, then you can construct a compact set $K$ satisfies for every $x \in G$, you have $K \cap xK \neq \phi$ see proposition 1.4.5 in the book of principles of Harmonic analysis of Deitmar and Echterhoof. Then you can easily show that $G=KK^{-1}$ then you can take $A=K$ and $B=K^{-1}$.

Abdo Elkasapy
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