If $A,B$ are Lebesgue measurable sets in $\mathbb{R}^n$ with $0 <\lambda(A),\lambda(B)< \infty$. Prove that $\lambda (\text{Int}(A+B))^{1/n} \ge \lambda(A)^{1/n}+ \lambda(B)^{1/n}$ where $\text{Int}(X)$ denotes the interior of a set $X$.
I have managed to prove that $A+B$ has non-empty interior using an argument very similar to the Steinhaus Theorem, on the same lines as Gina's proof: "Sum" of positive measure set contains an open interval? (of which I only learned when I gived up on this question and set to find a solution online with virtually no success, just the bits I already managed to do) I did learn another way of proving it, using the convolution thing - Steinhaus theorem (sums version), but if I am not mistaken both proofs give very small open sets.
I have since argued that by regularity theorem we can assume the sets are compact, which makes $A+B$ compact as well. (Basically if we prove it for any epsilon approximations we know that interior is bigger than $(\lambda(A)-\epsilon)^{1/n}+ (\lambda(B)-\epsilon)^{1/n}$ so letting epsilon tend to zero we get the result).
My general idea would be to exploit the fact that arbitrarily small, but non zero measure sets sum has non-empty interior and hopefully combine it with compactness. But have had virtually no luck, you can't really kick out any open ball you find in the sum as set sums are weird and you would need to kick out a lot. Also looking at $X=A+B- \text{Int}(A+B)$ we know it has non-zero measure, or we are done by standard Brunn-Minkowski so another route I tried was to show we must be able to find an open subset of $X$ to reach a contradiction, we note that $X$ is compact. But here the problem is that the preimage of $X$ in $A$ and $B$ respectively can still have measure zero (as eg sum of two cantor sets (digit 1's and digit 2's) contains everything).
This is a homework question that I am really, really interested in finding a solution to I would appreciate hints or a full solution, links to preferably not too advanced relevant topics and in fact any help you might be able to give.
Apologies if I made a mistake in the format, this is my first post so hope you might be able to forgive.
Many thanks, Paul.
