# Haar Measure of Product of Compact Sets

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 $$\mu(E_1E_2). But this doesn't work because $$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.

• 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". Feb 26 at 17:19

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

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