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In solving this question, I found that s quicker proof is possible with below net characterization of l.s.c. functions. Could you have a check on my attempt?

Let $X$ be a topological space and $f:X \to \mathbb R \cup \{-\infty, +\infty\}$. For $x \in X$, let $\mathcal N(x)$ be the set of all open neighborhoods of $x$. Then the following statements are equivalent.

(a) For all $x \in X$, $$ f(x) \le \liminf _{y \to x} f (y) := \sup _{V \in \mathcal N (x)} \inf _{y \in V \setminus \{x\}} f(y). $$

(b) For all $x \in X$, if $\alpha \in \mathbb R$ and $(x_d)_{d\in D}$ is a net such that $x_d \to x$ and $f(x_d) \to \alpha$, then $f(x) \le \alpha$.

I posted my proof separately as an answer below, so I can accept my own answer and thus remove my question from unanswered list. If other people post answers, I will happily accept theirs.

Update: I have found that lower semi-continuity can also be characterized by

(c) For all $x \in X$, if $(x_d)_{d\in D}$ is a net such that $x_d \to x$, then $$ f(x) \le \liminf_d f(x_d) := \sup_{d\in D} \inf_{e\ge d} f(x_e). $$

The discussion of limsup and liminf of nets can be found here and here.

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1 Answers1

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(a) $\implies$ (b). Fix $\varepsilon>0$. There is $d_1 \in D$ such that $x_d \in V$ for all $d \ge d_1$. There is $d_2 \in D$ such that $f(x_d) < \alpha + \varepsilon$ for all $d \ge d_2$. There is $d_3 \in D$ such that $d_3 \ge d_1$ and $d_3 \ge d_2$. Hence $x_{d_3} \in V$ and $f(x_{d_3})< \alpha + \varepsilon$. So $$ f(x) \le \sup _{V \in \mathcal N (x)} \alpha + \varepsilon = \alpha + \varepsilon. $$ This inequality holds for any $\varepsilon >0$, so $f(x) \le \alpha$.

(b) $\implies$ (a). Let $\varepsilon > 0$ and $$ \alpha :=\sup _{V \in \mathcal N (x)} \inf _{y \in V \setminus \{x\}} f(y). $$

WLOG, we assume $\alpha < +\infty$. By AC, for each $V \in \mathcal N(x)$, we pick $x_V \in V\setminus \{x\}$ such that $$ \inf _{y \in V \setminus \{x\}} f(y) \le y_V:= f(x_V) < \varepsilon + \inf _{y \in V \setminus \{x\}} f(y). $$

Clearly, $x_V \to x$. Fix $V_0 \in \mathcal N(x)$. Then $y_{V_0} - \varepsilon \le y_V \le \varepsilon + \alpha$ for all $V \ge V_0$. Because the interval $[y_{V_0} - \varepsilon, \varepsilon + \alpha]$ is compact, there is a subnet $(y_{\varphi(d)})_{d\in D}$ and $\beta \in [y_{V_0} - \varepsilon, \varepsilon + \alpha]$ such that $y_{\varphi(d)} \to \beta$. On the other hand, $x_{\varphi(d)} \to x$. It follows that $f(x) \le \beta \le \varepsilon + \alpha$. Because this inequality holds for every $\varepsilon>0$, we get $f(x) \le \alpha$. This completes the proof.

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