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 $$\alpha'_n x + \beta'_n \to ax+b$$.

• Phil, could you cite the source you took the snapshot of? It's appreciable if you mention the source alongside the image/snap. Jan 13, 2023 at 7:41
• @User1865345 sure it's added now
– Phil
Jan 13, 2023 at 15:14
• Thanks for the citation. Jan 13, 2023 at 15:15
• Jan 13, 2023 at 16:46

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.