This is a follow up of this question, which contained a mistake, as very helpfully pointed out by Snoop. Hence, the corrected version.
Let $F$ be the CDF of a distribution, supported on $[-1,1]$. $F$ has a density $f>0$.
Let $F_n$ be a sequence of distributions converging weakly to $F$, with the following property. For each $n$, $F_n$ is supported on $\{t \in [-1,1]: |t|=\frac{k}{2^n}\;\text{for some}\;k \in \mathbb{N} \cup \{0\}\}$, i.e. it is finitely supported. Let $f_n(t)$ be the probability of $t$, under $F_n$.
Define the sequence of probability distributions $g_n$ as follows:
$$\widetilde{g}_n(t)=\begin{cases}f_n(t),\;\text{if}\;|t|>\frac{1}{n},\\ 0,\;\text{otherwise}\;\end{cases},\\ \\ g_n(t)=\frac{\widetilde{g}_n(t)}{\sum\limits_{|t'|>\frac{1}{n}}f_n(t')}$$
My question is: Does the probability distribution with probability of $t=g_n(t)$, converge (weakly) to $F$?
As you can see, in the sequence $g_n$, 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 did not edit the earlier question because I wanted Snoop's answer on it, which replies to the original question, to remain, as it is very instructive. However if this is considered a duplicate I will be happy to edit the previous one and delete this one.