4.22 Semicontinuity

Definition 4.22.1 (Semicontinuous). 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 4.22.2. 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. (1): For any $\alpha \in \real$,

\[\bracs{f > \alpha}= \begin{cases}\emptyset &\alpha \ge 1 \\ U &\alpha \in [0, 1) \\ X &\alpha < 0\end{cases}\]

(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

\[\bracs{f + g > a}\supset \bracs{f > f(x_0) - \eps}\cap \bracs{g > g(x_0) - \eps}\in \cn^{o}(x_{0})\]

As this holds for all $x_{0} \in \bracs{f + g > a}$, $\bracs{f + g > a}$ is open by Lemma 4.4.3.

(4): For any $a \in \real$,

\[\bracs{F > a}= \bigcup_{f \in \mathcal{F}}\bracs{f > a}\]

is open.

(5): By Proposition 14.2.3.$\square$

Proposition 4.22.3. Let $X$ be a LCH space and $f: X \to [0, \infty]$ be lower semicontinuous, then

\[f = \sup_{\substack{\phi \in C_c(X) \\ 0 \le \phi \le f}}\phi\]

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 $a \in (0, f(x))$,

\[f(x) = \sup_{\substack{\phi \in C_c(X) \\ 0 \le \phi \le f}}\phi(x)\]
$\square$