# Trying to show a measure is inner regular on open sets

Background:

Let $X$ be a locally compact Hausdorff space, $C_{00}$ the collection of continuous functions on $X$ with relatively compact support. Then let $I \in C_{00}^*$ such that $f \in C_{00}$ nonnegative implies $I(f)\geq 0$.

Define for nonnegative lower semi continuous $g$, $$J(g) = sup(I(f): f\in C_{00}, 0\leq f \leq g)$$.

For general nonnegative $g$, $$K(g) = inf(J(f) : f\text{ lower semi continuous, }g \leq f)$$

My question:

Let $U \subset X$ be open. Let $b < K(\chi_U)$. Next line chooses a $f\in C_{00}, f\geq 0$ such that $I(f) > b$ and $f \leq \chi_U$. I cannot verify that this is possible from the definitions above. How can it be done?

-

$\chi_U$ is lower semicontinuous: Let $x \in X$ and $\epsilon > 0$. If $x \in U$, then $\chi_U$ is continuous at $x$ (as constant on the neighbourhood $U$ of $x$), so let $x \not\in U$. Then for each $y \in X$, we have $\chi_U(y) \ge \chi_U(x) - \epsilon = -\epsilon$. So $\chi_U$ is lower semicontinuous on $X \setminus U$ also.
Hence $K(\chi_U) = J(\chi_U)$ and the existence of $f$ follows from the definition of $J$.
I completely didn't see that $U$ being open made the characteristic function lower semi continuous. Thanks for pointing this out I feel silly. :) – roo Jun 9 '13 at 21:20
Actually. I guess $U$ doesn't even need to be open. – roo Jun 9 '13 at 21:26