I'm now studying Durret's probability theory and examples. In section 3.2, he discussed on 'vague convergence' of distribution functions.
He said a family of distribution functions $F_{n}(x)$ (i.e. non-decreasing, right-continuous and $F(\infty)=1$, $F(-\infty)=0$) vaguely converges to a function $F(x)$ if $F(x)$ is non-decreasing and right-continuous with $$F(x)=\lim_{n \to \infty}F_{n}(x)$$ for all continuity points $x$ of $F$.
In the above definition, I thought that the non-decreasing and right-continuous condition is necessary to define such a concept (i.e., well-definedness of the limit). However, I wonder if there is a limit of distribution functions which is not non-decreasing or right-continuous. By summarizing, the question that I want to ask is as follows.
Q. Let $F_{n}:\mathbb{R} \to [0,1]$ be a family of distribution functions and $F$ be any function. If $$\lim_{n \to \infty} F_{n}(x)=F(x)$$ for all continuity points $x$ of $F$, then is $F$ necessarily right-continuous? non-decreasing? If it is, how can I prove it? If not, is there any counter-example?
This might be a stupid question, but I got a little confused from this question.
Thank you for answering this question.
The limit $F$ is non-decreasing : if $x\leq y$, then for all $n\in\mathbb N$, $F_n(x)\leq F_n(y)$, and by taking the limit we get $F(x) \leq F(y)$.
However, it is not necessarily right-continuous and $F(+\infty) = 1,F(-\infty) =0$ do not hold automatically.
Let $\chi$ the distribution function for the uniform law on $[0,1]$, ie $$\chi(x) =\left\{\begin{array}{cl} 0 & \text{if } x\leq 0 \\ x & \text{if }x\in[0,1]\\ 1 &\text{if } x\geq 1\end{array}\right.$$
Then, with $F_n(x) = \chi(\frac xn)$, we have: $$\lim_{n\to +\infty} F_n(x) = \left\{ \begin{array}{cl}0 & \text{if } x\leq 0 \\1 & \text{if } x>0\end{array} \right.$$ which is not right-continuous.
With $F_n(x) = \chi(x +n)$, we have $\lim_{n\to +\infty} F_n(x) =1$ so $F(-\infty) \neq 0$.
With $F_n(x) = \chi(x-n)$, we have $\lim_{n\to +\infty} F_n(x) = 0$ so $F(+\infty) \neq 1$.