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.
 A: 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, $0= I_{(1,2)}(2) \neq lim_{k \to \infty}I_{(1,2+1/k)}(2)=1$.
A: 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 $$\lim_{n\to\infty} \mathsf 1_{(a,x_n)} = \lim_{n\to\infty}\mathsf 1_{(a,b_n)} = \mathsf 1_{(a,x_0)}.$$
A: If $a \neq x_0$, then the functions converge pointwise, but not uniformly. If $a = x_0$, then it is not guaranteed that the functions will converge unless the sequence is monotonic.
Let's just look at the unequal case, and without loss of generality let's assume that $a \lt x_0$. Then consider some point $z \in (a, x_0)$. It's clear that $1_{(a, x_0)}(z) = 1$, so we want to prove that $1_{(a, x_n)}(z) \rightarrow 1$. We know that $x_n \rightarrow x_0$, but let's be more formal:
$\forall(\epsilon > 0)\ \exists (n \in \mathbb{N})\ \forall (m > n)\ |x_m - x_0| \lt \epsilon$
i.e. for any given distance $\epsilon$, we can choose an integer $n$ such that all of the values of the sequence from $x_n$ onwards are closer to $x_0$ than $\epsilon$. So, let's choose $\epsilon = \frac{x_0 - z}{2}$. Then for all values of the sequence sufficiently large, $x_m$ is less than that distance to $x_0$, meaning that they lie between $z$ and $x_0 + \epsilon$, which means that $z$ is guaranteed to be in the interval $(a, x_m)$ and hence $1_{(a, x_m)}(z) = 1$. Similarly, all points to the right of $x_0 + \epsilon$ will lie outside the interval and so will return a value of 0 for the indicator function. Hence the sequence of functions converges pointwise.
