Semicontinuity problem If $A \subset \mathbb{R}^n$, is that claim true?
$$\chi_A \text{ is LSC} \Longleftrightarrow A\text{ is open}$$
And then how can I prove it?
($\chi_A$ is characteristic function : if $x \in A$ than $\chi_A =1$ otherwise zero.)
 A: True. In fact 
\begin{equation}
\chi_{A} \mbox{is LSC} \  \Leftrightarrow \ \forall \  t \in \mathbb{R} \  \{x : \chi_A (x) \le t\}\ \mbox{is closed.}
\end{equation} 
 But 
$$
\{x : \chi_A (x) \le t\} = \left\{
\begin{array}{rcl}
\mathbb{R^{n}} &if& t \ge 1\\
A^{c} &if& 0 \le t <1\\
\emptyset &if& 1 < t.\\
\end{array}
\right.
$$
$\Leftrightarrow A$ is open.
A: If by LSC you mean lower semicontinuity and by charcteristic function you mean $\chi_A(x) = \infty$ when $x \notin A$, in the convention of convex analysis, then, this does not hold. The set would be closed iff its characteristic function is LSC.
In general, the following properties are equivalent for an extended real-valued function:


*

*$f$ is LSC on $\mathbb{R}^n$,

*the epigraph of $f$ is closed in $\mathbb{R}^n \times \mathbb{R}$, 

*the sublevel sets $\text{lev}_{\le \alpha} f := \{ x \in \mathbb{R}^n: f(x) \le \alpha\}$ are all closed in $\mathbb{R}^n$.


If by $\chi_A$ you mean $\chi_A(x) = 0$ if $x \notin A$. Then, it holds. The sublevel sets of $\chi_A$ are $\mathbb{R}^n$, $\mathbb{R}^n \setminus A$ and the empty set. For a nonempty set $A$, all of these are closed iff $A$ is open. Hence, the claim is true.
