Let $U \subset \mathbb{R}$ be open. For a positive $\mathcal{B}(U)$-measurable function $u$, we define
\begin{eqnarray*} S_{u}:=\left\{ x \in U: \int_{ \overline{B_{\epsilon}(x)}} u^{-1}(t)dt < \infty \,\,{\rm for\,some }\,\, \epsilon>0 \right\} \end{eqnarray*}
where $\overline{B_{\epsilon}(x)}=\{y \in {U}:|y-x|\leq\epsilon\}$.
I want to show that $S_{u}$ is open and that $u>0\, \mu-a.e.$ on $S_{u}$ ($\mu$ is Lebesgue measure on $U$ )
The former is easy. In fact, let $x \in S_{u}$ then $B_{\frac{\epsilon}{2}}(x) \subset S_{u}$.
The latter: Since for any $x \in S_{u}$, $\int_{ \overline{B_{\epsilon}(x)}} u^{-1}(t)dt < \infty$ then
\begin{eqnarray*} \mu \left(t \in \overline{B_{\epsilon}(x)}: u(t)=0 \right)=0 \end{eqnarray*}
This implies $u(t)>0\,\mu-a.e. {\rm on}\,B_{\epsilon}(x) $. But my purpose is to prove
\begin{eqnarray*} \mu \left(t \in S_{u}: u(t)=0 \right)=0\tag1 \end{eqnarray*}
How do I get to $(1)$? Thanks.
Here is one approach, not necessarily shortest but it works. Write $U$ as a countable union of compact sets $K_n$, for example $K_n$ could consist of all points $x$ such that $|x|\le n$ and $\operatorname{dist}(x,\partial U)\ge 1/n$. This is called exhaustion by compact subsets.
Cover $K_n$ by balls of the form $B_\epsilon$ on which you know $u>0$ a.e. Take a finite subcover. Conclude $u>0$ a.e. on $K_n$. Take union over $n$.