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.

The answer is no. Let  $a=1$, $x_0=2$ and $x_k=2+1/k$. Then $I_{(1,2+1/k)} $ is not convergent to $ I_{(1,2)}$. Indeed, $0= I_{(1,2)}(2) \neq lim_{k \to \infty}I_{(1,2+1/k)}(2)=1$.
