# Convergence of generalized inverses

I copy-paste the post from Math Overflow - maybe someone can give me a tip.

During the reading about Fisher–Tippett–Gnedenko theorem (it can be found easily on wiki - I don't have enough reputation to post more links), I've got stuck, trying to understand more deeply one of the lemmas which lead to the final result. Instead of using CDFs in explanation why the distribution of normalized maxima can converge to one of the three distributions (Gumbel, Weibull or Frechet), many authors (de Haan, Resnick) do the same thing with the help of quantile functions. For that reason, it seems useful to prove this kind of lemma (here scan from De Haan L., Ferreira A., Extreme Value Theory An Introduction, p. 5; simmilar proposition - without proof - can be found here as well: http://cims.nyu.edu/~nica/Extreme_Values_Resnick.pdf)

As usual, wording "the proof of the left-hand inequality is similar" makes me a bit suspicious. As I understand correctly, remaining part should be started with choosing $0 < \epsilon_{1} < \epsilon$ such that $g^{\leftarrow}(x) + \epsilon_{1}$ is a continuity point of $g$. Of course, it can be done because of monotonicity of $g$. However, it should be also provided that $g^{\leftarrow}(x) + \epsilon_{1} \in (a,b)$, knowing that $x \in (g(a), g(b))$, which is not so obvious. It's easy to show the problem in a more graphical way (image below quoting text in the link). $u$ is of course a continuity point of $g^{\leftarrow}$, and of course for all $\epsilon_{1}$, $g^{\leftarrow}(u) + \epsilon_{1}$ is a continuity point of $g$, but there is no chance that $g^{\leftarrow}(x) + \epsilon_{1} \in (a,b)$. I am wondering, first of all, if this lemma in such a shape is valid (alternatively, how it can be refined) and, what is maybe more important/interesting, if convergence of inverses implies convergence: $$f_{n} \rightarrow g$$.

Because this lemma is useful in context of probability theory, I can assume that both $f_{n}$ and $g$ are CDFs (I don't know if this helps).

Thanks for any feedback.