I'm having trouble with the inequality in the proof of the following claim:
Let $E$ be a locally-compact separable metric space and let $\{\mu_n\}$ weak-* converge to $\mu$. If $\{|\mu_n|\}$ locally weak-* converges to $\lambda$ then $\lambda \geq |\mu|$
The proof begins:
Let $A \subset E$ be a relatively compact open set, define: $$A_t := \{x \in A | \text{dist}(x,\partial A) > t\}$$ and let $u \in C_c(A)$ be such that $\chi_{A_t} \leq u \leq \chi_A$. Then by Corollary 1.13 we have: $$|\mu|(A_t) \leq \liminf_{n \rightarrow \infty} |\mu_n|(A_t) \quad (*)$$
Corollary 1.13 reads:
Let $\mu:\mathcal{B}(E) \rightarrow [0,\infty]$ and $B \in \mathcal{B}(E)$. Then $$\mu(B) = \sup \{\mu(C)|C \subset B, C \text{ closed}\}$$ provided $\mu$ is $\sigma$ finite. Further: $$\mu(B) = \inf \{\mu(A)|A \supset B, A \text{ open}\}$$ holds if $E$ is a countable union of open sets, each of finite measure.
I'm confused about how they are obtaining the inequality. Are they even using the corollary? (they may not be, there are more steps in the proof -- but I didn't include them because it appeared to not be used in those steps).