limit of probability distribution Let $(F_n)$ and $F$ be one-dimensional cumulative distribution functions, with $F_n\to F$ in distribution. This means for all continuouity points $x$ of $F$ $\lim_{n\to\infty} F_n(x)=F(x)$ holds.
For real sequences $a_n\to a$ and $b_n\to b$ and all continuouity points $x$ of $F(ax+b)$ the following holds:
$$\lim_{n\to\infty} F_n(a_n x+b_n)=F(ax+b)$$
Intuitively this sounds reasonable, but I don't know how to show this mathematically.
I think, since distribution functions are continuous from the right, we have $\lim_{m\to\infty} F_n(a_m x+b_m)=F_n(ax+b)$, but I'm not sure how to go on with this.
 A: Let's take $a>0$. Then $G\left(x\right)=F\left(ax+b\right)$ and $G_{n}\left(x\right)=F_{n}\left(ax+b\right)$ are cdf's with 
$G_{n}\rightarrow G$ in distribution as a consequence of $F_{n}\rightarrow F$
in distribution.
Fix $x$.
For $\varepsilon>0$ you can find a continuity point $c$ of $G$ with
$x-\varepsilon <c<x$. This because the set of
its continuity points is dense. Note that $ac+b<a_{n}x+b_n$ for $n$ large enough, so:
$F\left(a\left(x-\varepsilon\right)+b\right)\leq F\left(ac+b\right)=\lim_{n\rightarrow\infty}F_{n}\left(ac+b\right)\leq\liminf_{n\rightarrow\infty}F_{n}\left(a_{n}x+b_{n}\right)$
Likewise $\limsup_{n\rightarrow\infty}F_{n}\left(a_{n}x+b_{n}\right)\leq F\left(a\left(x+\varepsilon\right)+b\right)$. 
Now if $x$ is a continuity point of $G$ then $\varepsilon\downarrow0$ results in: 
$F\left(ax+b\right)\leq\liminf_{n\rightarrow\infty}F_{n}\left(a_{n}x+b_{n}\right)\leq\limsup_{n\rightarrow\infty}F_{n}\left(a_{n}x+b_{n}\right)\leq F\left(ax+b\right)$
This shows that in that case $F_{n}\left(a_{n}x+b_{n}\right)$ converges to $F\left(ax+b\right)$.
