Question in real analysis an Minkowski difference

Question

I have the following question in real analysis: I was first asked to prove that if the function which I proved to be continuous

has a point x such that F(x) > 0 then there exists an open cube in A-B (this is the Minkowski difference of the two sets not set subtraction). This I could do but the last part puzzles me as I am asked to prove simply given A,B measurable with m(A),m(B) strictly positive with finite measure then there exists an open cube Q in A-B. Really need any help on this Thank you all P.S. all measure and integration in the sense of Lebesgue

Answer 1

If $F(x) > 0$, then $\chi_A(t)\chi_B(t-x) > 0$ on a set of positive measure. In particular, there exists at least one number $t$ with the property that $t \in A$ and $t-x \in B$. It follows that $x \in A-B$.

Since $F$ is continuous it is positive on a cube $Q$ containing $x$. For all $y \in Q$ the above argument shows $y \in A-B$. Thus $Q \subset A-B$.