Let $s\mapsto A_s$ be an increasing right continuous function, $s\geq 0$. Then define $A_s^{-1}:=\inf\{t\geq 0: A_t>s\}$ with the convention $\inf\{\emptyset\}=\infty$. By definition the map $s\mapsto A_s^{-1}$ in increasing so one can define $A_{s-}^{-1}:=\lim_{h\rightarrow 0} A_{s-h}^{-1}$ the left limit. I want to show that $A_{s-}^{-1}=\inf\{t\geq 0: A_t\geq s\}$
Unfortunately I don't see how to do this, I thought about rewriting $\inf\{t\geq 0: A_t\geq s\}=A_s^{-1}\wedge \inf\{t\geq 0: A_t=s\}$ but this does not help me. Then I remarked that $A_{s-h}^{-1}=\inf\{ t\geq 0: A_t+h>s\}\leq\inf\{t\geq 0: A_t\geq s\}$ hence I would say $lim_{h\rightarrow 0} A_{s-h}^{-1}\leq \inf\{t\geq 0: A_t\geq s\}$ but the other inequality is still not clear. Can someone help me?
Since $A_t$ is right continuous and increasing it is upper semi continuous. This means that $\{t\ge 0:A_t<s\}$ is open and $\{t\ge 0:A_t\ge s\}$ is closed. For any sequence $s_n\uparrow s$ we have $$ \bigcap_{n\in\mathbb N}\{t\ge 0:A_t> s_n\}=\{t\ge 0:A_t\ge s\}\, $$ and also $$ \bigcap_{n\in\mathbb N}\{t\ge 0:A_t\color{red}{\ge} s_n\}=\{t\ge 0:A_t\ge s\}\,. $$ The second relation holds because we can assume wlog that the sequence $s_n$ is strictly increasing. Since $$ A^{-1}_{s-}=\sup_{n\in \mathbb N}\inf\{t\ge 0:A_t> s_n\} $$ we must show the claim $\stackrel{!}{=}$ in \begin{align} \inf\{t\ge 0:A_t\ge s\}&= \inf\bigcap_{n\in\mathbb N}\{t\ge 0:A_t> s_n\}\\ &=\inf\bigcap_{n\in\mathbb N}\{t\ge 0:A_t\color{red}{\ge}s_n\}\\ &\stackrel{!}{=}\sup_{n\in \mathbb N}\inf\{t\ge 0:A_t> s_n\}\,.\tag{1} \end{align} Proof. Since $$ \bigcap_{n\in\mathbb N}\{t\ge 0:A_t> s_n\} \subset\{t\ge 0:A_t> s_m\} $$ holds for all $m$ we have $$ \inf\bigcap_{n\in\mathbb N}\{t\ge 0:A_t> s_n\} \ge\inf\{t\ge 0:A_t> s_m\} $$ so that the inequality $\ge$ in (1) follows. For the other inequality we observe that $$ \inf\{t\ge 0:A_t\color{red}{\ge}s_n\}\le \inf\{t\ge 0:A_t>s_n\} $$ so that it is enough to show $$\tag{2} \inf\bigcap_{n\in\mathbb N}\{t\ge 0:A_t\color{red}{\ge}s_n\} \le \sup_{n\in\mathbb N}\inf\{t\ge 0:A_t\color{red}{\ge}s_n\}\,. $$ Since the sets $\{t\ge 0:A_t\color{red}\ge s_n\}$ are closed every infimum of such a set is in the set. Then
$\sup\limits_{n\in\mathbb N}\inf\{t\ge 0:A_t\color{red}{\ge}s_n\}$ is in every $$ \inf\{t\ge 0:A_t\color{red}{\ge}s_m\} $$ since that set is closed for every $m\,.$ Therefore, that supremum is in the intersection of all those sets. It follows that this supremum must be larger or equal than the infimum on the LHS of (2).