Let $f : \mathbb R^n \to [-\infty, \infty]$ convex and let $f(\overline x) > -\infty$ for $\overline x \in \mbox{int}(\mbox{dom}(f)$. Show that $f(x) > -\infty$ for all $x \in \mathbb R$.
The involved definitions could be found here. I have conceptual difficulty with this exercise, because I read somewhere that convex functions are continuous, so intuitively with a jump from a value $x \in \mathbb R$ to some value $\pm \infty$, this continuity might be violated, or could the extended real line $[-\infty, \infty]$ incorporated with some suitable topology on it, such that continuity is still valid. Or does it simply mean that continuity is just supposed on $\mbox{dom}(f) = \{ x \in \mathbb R : f(x) \ne \pm \infty \}?$ I am asking because I guess continuity issues might play a role in solving this exercise.
To get a true statement, I think we need to add the assumption that $\text{dom} f$ has a non-empty interior.
Let $y \in \text{int}(\text{dom} f )$ and suppose (for a contradiction) that $f(x) = -\infty$ for some $x \in \mathbb R^n$. Convexity implies that $f(z) =- \infty$ for all points $z$ between $x$ and $y$. But some of the points between $x$ and $y$ belong to $\text{dom} f $, which is a contradiction.
Here is the last step in more detail. There exists $\epsilon > 0$ such that the closed ball of radius $\epsilon$ centered at $y$ is a subset of $\text{dom} f $. Choose $0 < t \leq \epsilon$ small enough that $z = y + t(x - y)$ is between $x$ and $y$. Then $f(z) = -\infty$, yet $z \in \text{dom} f$, which is a contradiction.