A finite continuous convex summand preserves proper closed convexity

Since $\mathrm{dom}g\mathrm{\ne }\mathrm{\varnothing }$, choose $x_0$ with $g(x_0)<+\mathrm{\infty }$. The value $f(x_0)$ is real, so $F(x_0)<+\mathrm{\infty }$ and $F$ is proper.

For $x,y$ with finite $F$ and $0\le \theta \le 1$, both $g(x),g(y)$ are finite. Applying convexity separately to $f$ and $g$ and adding gives $F((1-\theta )x+\theta y)\le (1-\theta )F(x)+\theta F(y)$; if either value of $F$ is $+\mathrm{\infty }$, the extended-real convexity inequality is automatic. Thus $F$ is convex.

For $r\in \mathbb{R}$, let $\mathrm{\Phi }(x,s)=(x,s-f(x))$. It is continuous: the first coordinates are unchanged, while continuity of $f$ makes the last coordinate $s-f(x)$ continuous. The identity $\mathrm{epi}F={\mathrm{\Phi }}^{-1}(\mathrm{epi}g)$ follows by rearranging $f(x)+g(x)\le s$, including points where $g(x)=+\mathrm{\infty }$. If a point lies outside $\mathrm{epi}F$, its image lies outside the closed set $\mathrm{epi}g$, so an open ball about that image misses $\mathrm{epi}g$; continuity of $\mathrm{\Phi }$ supplies a neighborhood of the original point mapped into that ball. Hence the complement of $\mathrm{epi}F$ is open and $F$ is closed.