Let $F$ be the CDF of a distrubtion, supported on $[-1,1]$. It has a density $f$ (See footnote 1). Consider the sequence of finitely supported distributions defined as follows.
Take $t\in [-1,1]$, $n \in \mathbb{N}$. Define:
$$\widetilde{f}_n(t)=\begin{cases}f(t),\;\text{if}\;|t|=\frac{1}{n}+\frac{k}{2^n}\;\text{for some}\;k \in \mathbb{N} \cup \{0\},\\ 0,\;\text{otherwise}\;\end{cases},\\ \\ f_n(t)=\frac{\widetilde{f}_n(t)}{\sum\limits_{\widetilde{f}_n(t')>0}\widetilde{f}_n(t')}$$
My question is: Does the probability distribution with probability of $t=f_n(t)$, converge (weakly) to $F$?
As you can see, in the sequence $f_n$, the distance between points in the positive or negative part of the support of $f_n$ $(1/2^n)$, reduces at a rate faster than the distance between the positive and negative parts of the support $(2/n)$.
I think the answer is yes, using the standard definition of distributional convergence as pointwise convergence of the CDF (in this case, because all points are points of continuity). But I still wanted to double check what exactly, if anything, we need to be careful about, regarding the differing rates of convergence in different parts of the support.
(Edit) Footnote 1: We can assume $f(t)>0$ for all $t \in [-1,1]$. I did not edit the question as it was, because it contained a mistake as pointed out by Snoop in the first answer. However I would like to allow ourselves to make the aforementioned simplifying assumption so we can focus on the role played by the differing rates of convergence.
The answer seems to be no for me. Let $f(t)=g(t)\mathbb{1}_{(\mathbb{R}/\mathbb{Q})\cap [-1,1]}(t)$ for some Lebesgue density $g$ on $[-1,1]$ with cdf $G(t)=\int_{[-1,t]}g(s)\lambda(ds)$. Then $F(t)=\int_{[-1,t]}f(s)\lambda(ds)=\int_{[-1,t]}g(s)\lambda(ds)=G(t)$. However, $f_n(t)=0/0$ for all $n$ and $t \in [-1,1]$.