5.22 Semicontinuity
Definition 5.22.1 (Semicontinuous).label Let $X$ be a topological space, $f: X \to (-\infty, \infty]$, and $g: X \to [-\infty, \infty)$, then $f$ is lower semicontinuous if for each $a \in \real$, $\bracs{f > \alpha}$ is open, and $g$ is upper semicontinuous if for each $a \in \real$, $\bracs{f < \alpha}$ is open.
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 $\alpha \in (-\infty, \infty)$,
which is closed. Therefore $f$ is lower semicontinuous.$\square$
Proposition 5.22.3.label Let $X$ be a topological space, then
- (1)
For any $U \subset X$ open, $\one_{U}$ is lower semicontinuous.
- (2)
For any $f: X \to (-\infty, \infty]$ lower semicontinuous and $\alpha \ge 0$, $\alpha f$ is lower semicontinuous.
- (3)
For any $f, g: X \to (-\infty, \infty]$ lower semicontinuous, $f + g$ is lower semicontinuous.
- (4)
For any collection $\mathcal{F}\subset (-\infty, \infty]^{X}$ of lower semicontinuous functions, $F = \sup_{f \in F}f$ is lower semicontinuous.
- (5)
For any $f: X \to (-\infty, \infty]$ lower semicontinuous, $f$ is Borel measurable.
Proof, [Proposition 7.11, Fol99]. (1): For any $\alpha \in \real$,
(2): If $\alpha = 0$, then $\alpha f = 0$ is continuous. If $\alpha > 0$, then for any $a \in \real$, $\bracs{\alpha f > a}= \bracs{f > a/\alpha}$ is open.
(3): Let $a \in \real$, $x_{0} \in \bracs{f + g > a}$, and $\eps \in (0, ((f + g)(x_{0}) - a)/2)$, then
As this holds for all $x_{0} \in \bracs{f + g > a}$, $\bracs{f + g > a}$ is open by Lemma 5.4.3.
(4): For any $a \in \real$,
is open.
(5): By Proposition 21.2.3.$\square$
Proposition 5.22.4.label Let $X$ be a LCH space and $f: X \to [0, \infty]$ be lower semicontinuous, then
Proof. Let $x \in X$ such that $f(x) > 0$ and $a \in (0, f(x))$, then $\bracs{f > a}$ is open. By Urysohn’s lemma, there exists $\phi \in C_{c}(X; [0, a])$ such that $\phi(x) = a$ and $\supp{\phi}\subset \bracs{f > a}$. As this holds for all $a \in (0, f(x))$,
$\square$
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