In the proof that I'm currently reading, there's a step that makes me quite confused:
"Hence $\hat{\mu_e} \in L^1(\mathbb{R})$ for $\sigma^{n-1}$ almost all $e \in S^{n-1}$ and therefore $\mu_e$ is a continuous function for such e, i.e., for all $\phi$, we have $\int g\phi d\mathcal{L}^1 = \int \phi d\mu_e$ for such $e$ where $g$ is continuous. As $\mu_e \in \mathcal{M}(A)$, we conclude that the interior of $A$ is non-empty for $\sigma^{n-1}$ almost all $e \in S^{n-1}$."
Context: $\mu_e \in \mathcal{M}(A)$ means that $\mu$ has a compact support $\subset A$ such that $0 < \mu_e(A) < \infty$
Any help would be greatly appreciated since I'm stuck here, thanks.
First off, we know that $\mu_{e}(A) > 0$. Thus, the associated function $g$ is not identically equal to $0$. Since $g$ is continuous, there is an open ball on which $g$ is nonzero. Finally, $\mu_{e}$ having compact support in $A$ means that the aforementioned open ball is contained in $A$.