Proposition 2.6.2132090FCDF4A

Proof. If $f\in C_ c(\Omega )$, $0\leq f\leq 1$ and $\operatorname{Supp}f\subset U$, then $\int f\, \mathrm d\mu =\lim _ i\int f\, \mathrm d\mu _ i\leq \varliminf _ i\mu _ i(U)$; taking the supremum over such $f$ and using (2.3.4) gives the first inequality. For the second, let $\varepsilon >0$, choose by outer regularity an open set $V\supset K$ with $\mu (V)<\mu (K)+\varepsilon $, and let $\eta $ be a cutoff function for $K$ in $V$. Then $\mu _ i(K)\leq \int \eta \, \mathrm d\mu _ i\to \int \eta \, \mathrm d\mu \leq \mu (V)$, so $\varlimsup _ i\mu _ i(K)\leq \mu (K)+\varepsilon $. Finally, for $E$ as in the last assertion, the closure $\overline E$ is a compact subset of $\Omega $, and $\overline E\setminus \operatorname{Int}E=\partial E$ is $\mu $-null, so $\mu (\operatorname{Int}E)=\mu (E)=\mu (\overline E)$. Since $\mu _ i(\operatorname{Int}E)\leq \mu _ i(E)\leq \mu _ i(\overline E)$, the two inequalities, applied to $\operatorname{Int}E$ and $\overline E$, give