Existence of regular Borel measure Let $X$ be a $\sigma$-compact and locally compact space, and let $\Lambda:C(X)\rightarrow \mathbb{C}$ be a linear functional such that $\Lambda(f)\ge0$ if $f\ge0$. How to show that exist exactly one regular Borel measure $\mu$ with compact support such that $\Lambda(f)=\int_{X}fd\mu$ for every $f\in C(X)$ ?
 A: Here's an idea, without all of the details.

Claim: Suppose that $f_n:X \to [0,\infty)$, is a sequence of continuous, positive functions such that, for each $x \in X$, there exists a neighbourhood of $x$ in which all but finitely many $f_n$ are identically zero. Then, there exists a number $N$ such that $n > N$ implies $\Lambda(f_n) = 0$.
Proof: By the finiteness assumption, the sum $S = \sum f_n$ makes sense and is another continuous, positive function $X \to [0,\infty)$. Note that, for each $n$, we have $S \geq S_n$ where $S_n$ is the $n$th partial sum of the series. So, by the positivity of the functional $\Lambda$, we have $\Lambda(S) \geq \Lambda(S_n) = \sum_n \Lambda(f_n)$ which, since the terms of the latter series are positive, implies that $\lim \Lambda(f_n)=0$. In fact, I claim that there is an $N$ such that $n > N$ implies $\Lambda(f_n) = 0$. Indeed, if this were not the case, then we could multiply the functions $f_n$ by a sufficiently fast-growing sequence of constants $C_n>0$ in order to obtain a new sequence of functions $C_n f_n$, still satisfying the hypotheses of the claim, but failing to have $\lim_{n \to \infty} \Lambda(C_n f_n) = \lim_{n \to \infty} C_n \Lambda(f_n) = 0$, contradicting what was just proven.

Since $X$ is a $\sigma$-compact, locally compact Hausdorff space, one ought to be able to construct to countable partition of unity consisting of compactly-supported functions. An application of the claim should be enough to show the functional is "compactly-supported" in a sufficiently useful sense to finish the problem.
