If $F_n(x)$ converges in distribution to $G(x)$, and $x_n \to x$, then does $F_n(x_n)$ converges in distribution to $G(x)$? I'm having some trouble verifying the following line from Extremes and Related Properties of
Random Sequences and Processes by Leadbetter:

If (1.2.1)' and (1.2.3)' hold, then obviously so does (1.2.2)'  

With $F'_n(x)$ converging weakly to $G(x)$ and $\alpha_n' \to a$ and $\beta'_n \to b$, it's clear that $F'_n(ax+b) \to G(ax+b)$ weakly. However, why should $F_n'(\alpha'_n x+ \beta'_n)$ converge weakly to $G(ax+b)$? Since $F_n'$ may not be continuous, it seems we cannot guarantee the convergence even though $\alpha'_n x + \beta'_n \to ax+b$.

If $X_n \to X$ in distribution and $b_n \to b$, then we may pick $Y_n = X_n$ in distribution for $1 \leq n \leq \infty$ with $Y_n \to Y$ almost surely.
Clearly ,$Y_n + b_n \to Y + b$ almost surely, so $X_n + b_n \to X+b$ in distribution.
We can prove the same thing for $a_n X_n \to aX$, which means $\frac{X_n - \alpha
'}{\beta_n'} \to \frac{X-a}{b}$ in distribution.
