I'm working my way through Compact convex sets and boundary integrals by Alfsen and during the proof of Proposition I.2.8 they use that set $W=\{\mu\in M_1^+(K)|\mu(\overline{V})<\alpha\}$ is vaguely open, where $K$ is compact Hausdorff space, $M_1^+(K)\subset M(K)=\left(C_{\mathbb{R}}(K)\right)^\ast$ are probability measures, $V$ is open subset of $K$ and $\alpha>0$. (If I understand correctly, vaguely open is the same thing as weak-* open, when $M(K)$ is viewed as a dual of $C_{\mathbb{R}}(K)$.)
They claim this set to be vaguely open, since
the function $\mu\mapsto\mu(\overline{V})=\mu(\chi_{\overline{V}})$ is vaguely u.s.c., since generally $\mu\mapsto \mu(f)$ is vaguely u.s.c. whenever $f$ is u.s.c.
I understand that $\chi_{\overline{V}}$ is u.s.c and why $\mu\mapsto \mu(\overline{V})$ being vaguely u.s.c. is enough, but I don't understand, why $f$ u.s.c. implies $\mu\mapsto\mu(f)$ being vaguely u.s.c. In the book all that is written as a explanation if
cf. (2.3)
where (2.3) is the formula $\mu(f)=\sup_\alpha\mu(f_\alpha)=\lim_\alpha\mu(f_\alpha)$ for $\mu\in M(K)$ positive measure, $f_\alpha$ increasing net in $C_{\mathbb{R}}(K)$ with $\sup_\alpha\mu(f_\alpha)<\infty$ and $f=\sup_\alpha f_\alpha$. I understand why this is true, but I have no idea how does it imply that $f$ u.s.c. implies $\mu\mapsto\mu(f)$ is vaguely u.s.c.