(continuation of the proof)
This series converges uniformly on $X$, so $g$ is continuous. Also the support of $g$ lies in $\overline{V}$. Since $2^{-n}h_n(x)=t_n(x)$ except in $V_n-K_n,$ we have $f(x)=g(x)$ except in $\cup(V_n-K_n).$
why is this true $2^{-n}h_n(x)=t_n(x)$ except in $V_n-K_n$?
why "we have $f(x)=g(x)$ except in $\cup(V_n-K_n)$ "
The argument for this is because if it holds for A holds for $\cup A$ always ?