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?



Identification of $\lim_{n\to\infty}f_n^{-1}$ with $f_n:\mathbb R_+\to (0,1]$ strictly decreasing and converging pointwise
$\newcommand{\de}{\delta}\newcommand{\ep}{\varepsilon}$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. $\quad\Box$


