According to the 7.3.5. Corollary of the book "Measure Theory" by V.I. Bogachev we have the following result:
Let $(X,\tau)$ be a completely regular space and let $\Gamma$ be a family of continuous function on $X$ separating the points in $X$. Denote by $\mathfrak{A}(\Gamma)$ the algebra on $X$ generated by the set $\{f^{-1}(U):f\in \Gamma\wedge U\subseteq\mathbb{R}\text{ is open}\}$. If $\mu:\mathfrak{A}(\Gamma)\to \mathbb{R}$ is outer regular and tight (w.r.t. the topology $\tau$) additive set function of bounded variation, then there's an unique Radon measure $\hat\mu :\mathfrak{B}_X\to \mathbb{R}$ that extends $\mu$.
My question is: In that proposition is it enough that $\mu$ is tight and outer regular w.r.t. the topology $\tau$?
Denote by $\sigma (X,\Gamma)$ the weak topology on $X$ generated by $\Gamma$ (coarsest topology on $X$ w.r.t. which all functions in $\Gamma$ is continuous).
Seeing the proof of that proposition, I think that, besides $\mu$ being tight w.r.t. $\tau$, it's necessary that $\mu$ is outer regular w.r.t. the topology $\sigma (X,\Gamma)$. That is, we need to show that for any $A\in \mathfrak{A}(\Gamma)$ and $\varepsilon >0$ there's a closed set $F$ in $(X,\sigma (X,\Gamma))$ such that $F\subseteq A$ and $|\mu |(A\setminus F)<\varepsilon $.
W.r.t. tightness, it's easy to show that if $\mu$ is tight w.r.t. to $\tau$, then $\mu$ is also tight w.r.t. to $\sigma (X,\Gamma)$. However, I don't know if this happen in the case of outer regularity.
