If $A$ and $B$ are measurable sets of $\mathbb{R}^n$ with strictly positive but finite measure, my problem is to prove that there is a vector $c \in \mathbb{R}^n$ such that $m((A+c) \cap B) > 0$.
I can show that outer measure (and therefore measure) is translation invariant, but I don't know how to get a vector $c$ such that $A+c$ and $B$ line up nicely.
This problem already has a solution here, but I'm trying to make an argument based on the outer measure rather than convolution.
My attempt:
The outer measure $m^*(A) = \inf \{ \sum v(Q_j)$ $|$ $Q_j$ are open $n$-cubes and $A \subseteq \cup Q_j \}$, and $m^*(B)$ is defined similarly. Let $C$ be an $n$-cube that contributes to $m^*(A)$ and let $D$ be an $n$-cube that contributes to $m^*(B)$.
Let $a \in C$ and $ b \in D$ whose coordinate values are lowest. Since these are points in $\mathbb{R}^n$, there is a vector $c$ such that $a + c = b$. Furthermore, $C+c$ intersects with $D$ in a nontrivial $n$-cube. Therefore the outer measure of $(A+c) \cap B$ is positive, and since measurability is translation invariant and closed under intersections, and the outer measure of a measurable set is its measure, $m((A+c) \cap B)$ is positive.
My main doubt here is that I'm not sure if this definition for outer measure is valid. In the general definition for outer measure, the $Q_j$s have to be measurable, not open. Open sets are $\epsilon$-close to measurable sets, but I don't have the intuition about analysis to say whether or not this is acceptable.
