Proposition 21.11.868114EF2F29C

Proof. For every $x\in N$ and $a\in\mathbb R$, the inequality $a\geq\sup_iF_i(x)$ holds exactly when $a\geq F_i(x)$ for every $i$. Consequently \[\operatorname{epi}(\sup_iF_i)=\bigcap_i\operatorname{epi}F_i.\] An intersection of convex sets is convex, so the supremum is convex. This also covers extended values and an empty epigraph.