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?


1 Answer 1


$\newcommand{\de}{\delta}\newcommand{\ep}{\varepsilon}$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)<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)\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)<x_0-\ep\implies f_n(x_0-\ep)<t_0 \implies f(x_0-\ep)<t_0+\de \\ & \implies x_0-\ep\ge f^{-1}(t_0+\de)>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. $\quad\Box$

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

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