Convergence of indicator functions Suppose $x_0 \in \mathbb{R}$ and $\{x_n\}_{n \geq 1}$ be a sequence converging to $x_0$, i.e, $x_n \rightarrow x_0.$ Is it true for a fixed $a\in\mathbb{R}$ that:
$$
1_{(a,x_n)} \rightarrow 1 _{(a,x_0)}
$$
where $1_{A}(y)=1$ if $y \in A$ and $0$ otherwise.
I've tried proving the same and it's not quite intuitive if $x_n$ is not a monotonic sequence. That's where I've been stuck.

If $x_n$ is a convergent sequence, then $$\lim_{n\to\infty} x_n = \liminf_{n\to\infty} x_n = \limsup_{n\to\infty} x_n.$$
Now
\begin{align}
\liminf_{n\to\infty} x_n &= \lim_{n\to\infty}\inf_{k\geqslant n} x_k\\
\limsup_{n\to\infty} x_n &= \lim_{n\to\infty}\sup_{k\geqslant n} x_k,
\end{align}
and $b_n:=\inf_{k\geqslant n}x_k$ is a monotone increasing sequence while $c_n:=\sup_{k\geqslant n}x_k$ is a monotone decreasing sequence. Assuming that $a<x_0$, then $\mathsf 1_{(a,b_n)}$ is monotone increasing so $$\lim_{n\to\infty} \mathsf 1_{(a,x_n)} = \lim_{n\to\infty}\mathsf 1_{(a,b_n)} = \mathsf 1_{(a,x_0)}.$$
