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 $K$, [L2] with $U=X$ gives a cutoff $0\le h\le 1$ equal to $1$ on $K$. Thus $\mu (K)\le \int hd\mu =L(h)\le \mathrm{\parallel }L\mathrm{\parallel }$. Inner regularity on the open set $X$ therefore yields $\mu (X)\le \mathrm{\parallel }L\mathrm{\parallel }<\mathrm{\infty }$. This is where Dependent Choice enters.

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 $f\in C_c(X)$. To see uniform density directly, for $f\in C_0(X)$ and $ϵ>0$ the set $K=\{\mathrm{\mid }f\mathrm{\mid }\ge ϵ\}$ is compact; [L2] with $U=X$ gives $h\in C_c(X)$ equal to one on $K$ and between zero and one, so $\mathrm{\parallel }f-hf{\mathrm{\parallel }}_{\mathrm{\infty }}\le ϵ$. Since $L$ is bounded 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.

Let $E$ be Borel and $ϵ>0$. Open inner regularity of $X$ gives a compact $H\subseteq X$ with $\mu (X\setminus H)<ϵ$. Outer regularity of $X\setminus E$ gives an open $V\supseteq X\setminus E$ with $\mu (V)<\mu (X\setminus E)+ϵ$. Then $K:=H\setminus V$ is compact and contained in $E$, while $E\setminus K\subseteq (X\setminus H)\cup (E\cap V)$ and $\mu (E\cap V)=\mu (V)-\mu (X\setminus E)<ϵ$, using finiteness of $\mu$. Thus $\mu (E)\le \mu (K)+2ϵ$; letting $ϵ\downarrow 0$ proves compact inner regularity for every Borel $E$. Uniqueness follows from [L1].