Open sets are Caratheodory measurable for the RMK outer measure

Fix open $G,V$ and $f\prec V\cap G$. Put [given] $K=\mathrm{supp}f$. For any $g\prec V\setminus K$, the supports of $f$ and $g$ are disjoint, so $f+g\prec V$. Hence $\rho (V)\ge \mathrm{\Lambda }(f)+\rho (V\setminus K)$. Since $V\setminus G\subseteq V\setminus K$, $\rho (V\setminus K)\ge {\mu }^{\ast }(V\setminus G)$. Taking the supremum over $f\prec V\cap G$ therefore gives $\rho (V)\ge \rho (V\cap G)+{\mu }^{\ast }(V\setminus G)={\mu }^{\ast }(V\cap G)+{\mu }^{\ast }(V\setminus G)\mathrm{.}$

For arbitrary $E$ and open $V\supseteq E$, monotonicity and step 1.1 [step 1.1] give $\rho (V)\ge {\mu }^{\ast }(E\cap G)+{\mu }^{\ast }(E\setminus G)\mathrm{.}$ Infimizing over $V$ gives the hard Caratheodory inequality; outer subadditivity gives the reverse inequality.

Therefore every open $G$ is Caratheodory measurable. By [L1], the measurable sets form a complete sigma-algebra; since they contain all opens, they contain $\mathcal{B}(X)$, and the restriction is a Borel measure.