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This is from Bachman's Harmonic Analysis book, exercise 8.3

Let $\mu$ be a Haar measure in $G$ and let $E_1$ and $E_2$ be two compact subsets of $G$ such that $\mu(E_1)=\mu(E_2)=0$. Does this imply that $\mu(E_1E_2)=0$?

My attempt: $E_1E_2$ is compact. By outer regularity, consider an open neighborhood $U_2$ of $E_2$ with $\mu(U_2)<\epsilon$. $\{xU_2\}_{x\in E_1}$ is an open cover of $E_1E_2$ with, by left invariance, $\mu(xU_2)=\epsilon$. Let the finite subcover be $\{x_iU_2\}_{i=1}^n$. So $\mu(E_1E_2)<n\epsilon$. But this doesn't work because $n$ is dependent on $\epsilon$. Also, I didn't use the compactness of $E_1$. I don't even know whether the hypothesis implies that $\mu(E_1E_2)=0$.

Any help is appreciated. Thanks.

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    $\begingroup$ Have you tried looking for a counterexample? (Hint!) Perhaps a helpful idea is that "the product of low-dimensional things can be high-dimensional" and "low-dimensional things have measure zero". $\endgroup$ Feb 26 at 17:19

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Thanks to @Izaak van Dongen's comment, the statement is false. A counterexample: $G=(\mathbb{R}^2,+)$ with the Euclidean topology. $E_1=[0, 1]\times\{0\}$, $E_2=\{0\}\times [0, 1]$.

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An example in $\mathbb R$: the Cantor set $C$ has measure zero and $C+C=[0,2]$. A proof is in this article.

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