This question is a doubt I had while reading this answer.
The answer poster, of the linked answer says that $\lim_n F_{X_n}(0)$ does not exist, where $F_{X_n}(\cdot)$ is defined as follows:
$F_{X_{n}}(t)=0\cdot\boldsymbol{1}_{(-\infty,0)}(t)+tn\cdot\boldsymbol{1}_{[0,1/n)}(t)+1\cdot\boldsymbol{1}_{[1/n,\infty)}(t) =\begin{cases} 0,~\text{if}~ -\infty< x <0\\ nx, ~\text{if}~ 0\leq x\leq 1/n\\ 1, ~\text{if} ~1/n\leq x\leq \infty \end{cases} $
But, since $F_{X_n}(0)=0, \forall n$, hence $\lim_n F_{X_n}(0)=0$, hence $\lim_n F_{X_n}(0)$ exists.
Did I do something wrong?
If yes then what is it, and if no, could you give an example of a sequence of functions $F_{n}$ such that $\lim_n F_{n}(x)$ does not exist for some $x$
Let $X$ take the values $0$ and $1$ with probabilty $\frac 1 2$ each. Let $X_n=X+\frac 1 n$ if $n$ is even and $X-\frac 1 n=$ if $n$ is odd. Then $X_n \to X$ in probability, $P(X_n \leq 0)=0$ for $n$ even and $P(X_n \leq 0)=\frac 1 2$ for $n$ odd. So $\lim P(X_n \leq 0)$ does not exist.