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Suppose $f:\mathbb{R} \rightarrow \mathbb{R}$ is an upper semicontinuous function (USC), i.e. for all $\alpha \in \mathbb{R}$, the set $\{ x \in \mathbb{R} : f(x) < \alpha \}$ is open.

What are the set of discontinuities of $f$? For $f = \chi_{[- \infty, 0)}$, its discontinuity is $x=0$. Hence, the set of discontinuities of an USC is finite. Do we have example such that set of discontinuities of an USC is countable or uncountable?

Here states that characteristic function of Cantor set has uncountable discontinuities. But there is no justification on why such function is upper semicontinuous.

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Since, for a set $S\subseteq \Bbb R$, $$\{x\in\Bbb R\,:\,\chi_S(x)<\alpha\}=\begin{cases}\Bbb R&\text{if }\alpha>1\\\Bbb R\setminus S&\text{if }0<\alpha\le1\\\emptyset&\text{if }\alpha<0\end{cases}$$

an indicator function is USC if and only if $S$ is closed, and LSC if and only if $S$ is open.

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