Pointwise supremum of a convex function collection In Hoang Tuy, Convex Analysis and Global Optimization, Kluwer, pag. 46, I read:
"A positive combination of finitely many proper convex functions on $R^n$ is convex. The upper envelope (pointwise supremum) of an arbitrary family of convex functions is convex". 
In order to prove the second claim the author sets the pointwise supremum as:
$$
f(x) = \sup \{f_i(x) \mid i \in I\} 
$$
Then 
$$
\mathrm{epi} f = \bigcap_{i \in I} \mathrm{epi} f_i 
$$
As  "the intersection of a family of convex sets is a convex set" the thesis follows.
The claim on the intersections raises some doubts for me. 
If $(x,t)$ is in the epigraph of $f$,   $f(x) \leq t)$ and, as  $f(x)\geq f_i(x)$, also  $f_i(x) \leq t)$; therefore $(x,t)$ is in the epigraph of every $f_i$ and the intersection proposition follows. 
Now, what if, for some (not all) $\hat{i}$, $f_{\hat{i}}(x^0)$ is not defined? The sup still applies to the other $f_i$ and so, in as far as the sup is finite, $f(x^0)$ is  defined. In this case when  $(x^0,t) \in  \mathrm{epi} f$ not $\in \mathrm{epi} f_{\hat{i}}$ too, since $x^0 \not \in \mathrm{dom}\, f_{\hat{i}}$.
So it is simple to say that $f$ is convex over $\bigcap_{i \in I} \mathrm{dom} f_i\subset \mathrm{dom} f$, because in this set every  $x  \in \mathrm{dom}\, f_{\hat{i}}$.
What can we say when   $x \in \mathrm{dom}\, f$ but $x \not \in \mathrm{dom}\, f_{\hat{i}}$ for some $i$?

I think it is either assumed that the $f_i$ are defined on the same domain $D$, or that (following a common convention) we set $f_i(x)=+\infty$ if $x \notin \mathrm{Dom}(f_i)$. You can easily check that under this convention, the extended $f_i$ still remain convex and the claim is true.
