Lower semicontinuity of indicator function For any set $\mathcal{S} \subseteq \mathbb{R}^{N}$, let us define the indicator function $$\delta_{\mathcal{S}}(\mathbf{x}) \triangleq \begin{cases}
0, & \quad \textrm{if } \mathbf{x} \in \mathcal{S} \\
\infty, & \quad \textrm{otherwise}.
\end{cases}$$
Is this function lower semicontinuous? I suppose it is if the set $\mathcal{S}$ is closed, but I'm not absolutely sure.
 A: You are right.
Lower semicontinuity is equivalent to:
$$
\forall x \ \ \ \{ y| \delta_S(y)\le x
\}
$$is closed. 
Let us prove this from OP's definition:
$\delta_S$ is lsc iff $\forall x\  \liminf_x  \delta_S = \delta_S(x)$


*

*if $\liminf_x  \delta_S = \delta_S(x)$, then let 
$(y_n)\in \{ y| \delta_S(y)\le x\}^{\Bbb N}$ a convergent sequence.
$$
\delta_S(y_n)\le x\\
$$and as $n\to\infty$:
$$
\delta_S(y) = \liminf \delta_S(y_n) \le x \\
$$
so $\{ y| \delta_S(y)\le x\}^{\Bbb N}$ is closed.

*if $\forall x \ \ \ \{ y| \delta_S(y)\le x\}$ is closed:
let $x_n \to x$ a sequence such as $\delta_S(x_n)\to  \liminf_x \delta_S $.
For $u$ close enough to $x\in \{ y| \delta_S(y)> \delta_S(x) - r\}$ which is an open set, 
$$ \delta_S(u)> \delta_S(x) - r
\\
\liminf \delta_S = \lim \delta_S(x_n) \ge  \delta_S(x) - r
$$
so $$
\liminf \delta_S\ge  \delta_S(x)
$$
and there is equality (the other inequality is always true).


Here this set is $\emptyset$ or $S$, so lower semicontinuity is equivalent to: $S$ is closed.
