Positive C_0(X) functionals have finite regular representing measures

Restrict $L$ to $C_c(X)$ and apply [L1], obtaining a Radon measure $\mu$. For every compact $K$, a cutoff $0\le h\le 1$ equal to $1$ on $K$ gives $\mu (K)\le L(h)\le \mathrm{\parallel }L\mathrm{\parallel }$. Inner regularity on $X$ therefore yields $\mu (X)\le \mathrm{\parallel }L\mathrm{\parallel }<\mathrm{\infty }$.

Conversely, $\mathrm{\mid }L(f)\mathrm{\mid }\le L(\mathrm{\mid }f\mathrm{\mid })\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mu (X)$ first for $f\in C_c(X)$. Since $C_c(X)$ is uniformly dense in $C_0(X)$ and $\mu$ is finite, both sides extend continuously to $C_0(X)$, giving the representation and $\mathrm{\parallel }L\mathrm{\parallel }\le \mu (X)$. Thus equality holds.

A finite Radon measure is compact-inner-regular on every Borel set: apply open inner regularity to an open superset of the complement and use finite complements. Hence $\mu$ is regular. Uniqueness follows from [L1].