Proposition 5.22.2.label Let $X$ be a topological space, $f: X \to (-\infty, \infty]$, then $f$ is lower semicontinuous if and only if $\text{epi}(f)$ is closed.
Proof. Suppose that $f$ is lower semicontinuous. Let $x \in X$ and $y \in (-\infty, \infty)$ with $y < f(x)$, and $\alpha \in (y, f(x))$, then $\bracs{f > \alpha}\in \cn^{o}_{A}(x)$, and hence $\bracs{f > \alpha}\times (-\infty, \alpha)$ is a neighbourhood of $(x, y)$ contained in the complement of $\text{epi}(f)$. Therefore $\text{epi}(f)$ is closed.
Now suppose that $\text{epi}(f)$ is closed, then for each $\alpha \in (-\infty, \infty)$,
\[\bracs{f \le \alpha}= \pi_{1}[\text{epi}(f) \cap A \times \bracs{\alpha}]\]
which is closed. Therefore $f$ is lower semicontinuous.$\square$
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