I was reading Introduction to Geometric Metric Theory by Leon Simon. Here's the full statement of the Riesz representation theorem for non-negative functionals.
Suppose $X$ is a locally compact Hausdorff space, $\lambda:\mathcal{K}_{+}\to[0,\infty)$ with $\lambda(cf)=c\lambda(f)$, and $\lambda(f+g)=\lambda(f)+\lambda(g)$ whenever $c>0$ and $f,g\in\mathcal{K}_{+}$, where $\mathcal{K}_{+}$ is the of all non-negative continuous functions $f$ on $X$ with compact support. Then there is a Radon measure $\mu$ such that $\lambda(f)=\int_{X}f\,d\mu$ for all $f\in\mathcal{K}_{+}$.
The link below is the beginning of the proof. I have some problems understanding the last line. I could only see that $\mu(K)\leq\inf\{\lambda(g):g\in\mathcal{K}_{+},g\leq 1,g=1\mbox{ in a nhd. of }K\}$ by the definition straightforward. But I can't prove the converse. Could somebody give me some advice?
Edit: $W$ in $(**)$ is just an open set contained in another open set $U$.
beginning of the proof
Seeking a contradiction, assume that $$\mu(K)<A:=\inf\{\lambda(g):g\in \mathcal{K}_+,g\leqslant 1,g\equiv 1\text{ in a nbhd of }K\}.$$ Fix $$\mu(K)<a<A.$$ Since $$a=\mu(K)=\inf\{\mu(U):U\text{ open},U\supset K\},$$ $a$ is not a lower bound for $\{\mu(U):U\text{ open},U\supset K\}$, and there exists an open set $U$ containing $K$ such that $\mu(U)<a$. By replacing $U$ with a subset thereof, we can assume without loss of generality that $\overline{U}$ is compact and that there exists an open set $V$ such that $K\subset V\subset \overline{V}\subset U$.
By Urysohn's Lemma, there exists $g:X\to [0,1]$ which is continuous such that $g|_K\equiv 1$ and $g|_{X\setminus \overline{V}}\equiv 0$. Then $g\in \mathcal{K}_+$, $g\leqslant 1$, and $\text{supp}(g)\subset U$. Then $$A\leqslant \lambda(g)\leqslant \mu(U)<a<A,$$ a contradiction.