# Identification of $\lim_{n\to\infty}f_n^{-1}$ with $f_n:\mathbb R_+\to (0,1]$ strictly decreasing and converging pointwise

Let $$f_n: \mathbb R_+\to (0,1]$$ be continuous and strictly decreasing for every $$n\ge 1$$. Assume that the pointwise limit of $$(f_n)_{n\ge 1}$$ exists, denoted by $$f$$, and is also strictly decreasing. Can we prove

$$\lim_{n\to\infty}f_n^{-1}(t)=f^{-1}(t), \quad \mbox{for almost every } t\in (0,1)?$$

The definition of $$f_n^{-1}$$ is standard and we set $$f_n^{-1}(t):=\infty$$ for all $$t\in [0, f_n(\infty)]$$. As for $$f$$, which generalized inverse we should take for the above purpose?

$$\newcommand{\de}{\delta}\newcommand{\ep}{\varepsilon}$$Yes, this is true. Moreover, the continuity and the strictness of the decrease of the $$f_n$$'s are not needed.

Indeed, for any nonincreasing function $$g\colon[0,\infty)\to(0,1]$$ and any $$t\in(0,1)$$, let $$\begin{equation*} g^{-1}(t):=\sup\{x\in[0,\infty)\colon g(x)\ge t\}, \tag{0}\label{0} \end{equation*}$$ with $$g^{-1}(t):=0$$ if $$\{x\in[0,\infty)\colon g(x)\ge t\}=\emptyset$$. Of course, in the particular case when the function $$g$$ is continuously and strictly decreasing from $$1$$ to $$0$$ on $$[0,\infty)$$, the generalized inverse $$g^{-1}$$ defined by formula \eqref{1} coincides with the usual inverse of $$g$$.

Let us show that $$\begin{equation*} f_n^{-1}\to f^{-1} \tag{1}\label{1} \end{equation*}$$ pointwise on the set $$\begin{equation*} C_*:=C\cup(0,t_\infty), \end{equation*}$$ where $$t_\infty:=\lim_{x\to\infty}f(x)\in[0,1]$$ and $$C$$ is the set of all points $$t\in(0,1)$$ such that $$f^{-1}$$ is finite and continuous at $$t$$. Note that $$f^{-1}=\infty$$ on $$(0,t_\infty)$$.

Since the function $$f^{-1}$$ is nonincreasing, the set $$(0,1)\setminus C_*$$ is at most countable. So, it will follow that the convergence \eqref{1} is almost everywhere (a.e.), as desired.

To prove \eqref{1}, take first any $$t_0\in C$$ and let $$x_0:=f^{-1}(t_0)$$, so that $$x_0\in[0,\infty)$$. Take any real $$\ep>0$$. The condition $$t_0\in C$$ implies that for some real $$\de=\de_\ep>0$$ we have $$\begin{equation*} |f^{-1}(t_0\pm\de)-f^{-1}(t_0)|<\ep. \end{equation*}$$ By the pointwise convergence $$f_n\to f$$, there is some natural $$n_\ep=n_{\ep,\de_\ep}$$ such that $$\begin{equation*} n\ge n_\ep\implies |f_n(x_0\pm\ep)-f(x_0\pm\ep)|<\de. \end{equation*}$$

So, for $$n\ge n_\ep$$ we have the following implications: \begin{equation*} \begin{aligned} &f_n^{-1}(t_0)>x_0+\ep\implies f_n(x_0+\ep)\ge t_0 \implies f(x_0+\ep)>t_0-\de \\ & \implies x_0+\ep\le f^{-1}(t_0-\de) a contradiction. So, for $$n\ge n_\ep$$,
$$\begin{equation*} f_n^{-1}(t_0)\le x_0+\ep=f^{-1}(t_0)+\ep. \end{equation*}$$ Similarly, for $$n\ge n_\ep$$ we have \begin{equation*} \begin{aligned} & f_n^{-1}(t_0)f^{-1}(t_0)-\ep=x_0-\ep, \end{aligned} \end{equation*} a contradiction. So, for $$n\ge n_\ep$$,
$$\begin{equation*} f_n^{-1}(t_0)\ge x_0-\ep=f^{-1}(t_0)-\ep. \end{equation*}$$ Thus, convergence \eqref{1} holds on $$C$$.

Take now any $$t_0\in(0,t_\infty)$$, so that $$f^{-1}(t_0)=\infty$$. Take any real $$x\ge0$$. Then $$f(x)>t_0$$. So, there is some natural $$n_x$$ such that for all $$n\ge n_x$$ we have $$f_n(x)>t_0$$ and hence $$f_n^{-1}(t_0)\ge x$$. It follows that $$f_n^{-1}(t_0)\to\infty=f^{-1}(t_0)$$. So, convergence \eqref{1} holds on $$(0,t_\infty)$$ as well, and thus it does hold on $$C_*$$, as claimed. $$\quad\Box$$

• Beautiful arguments. Thanks a lot for the answer
– user478492
Jul 13, 2022 at 6:49
• @Philo18 : Thank you for your appreciation. Jul 13, 2022 at 11:24