TITLE: Proof that the generalized inverse of an increasing right-continuous function is also right-continuous
QUESTION [7 upvotes]: Let $a:[0,\infty) \to [0,\infty]$ be an increasing right continuous function, and $\tau:[0,\infty) \to [0,\infty]$ be the generalized inverse, i.e. $\tau(s):=\inf\{t\ge 0: a(t)>s\}, \inf \emptyset = \infty$. Prove that $\tau$ is right continuous. 
Proof. $$\{t:a(t)>s\} = \inf_{\epsilon>0} \{t:a(t)>s+\epsilon\}.$$ Therefore, $\inf\{t\ge 0: a(t)>s\} = \inf_{\epsilon>0} \inf \{t\ge 0: a(t)>s+\epsilon\}$ proving right continuity. 
I have trouble figuring out why this last equality proves right-continuity. Is there a way to see this immediately? I would greatly appreciate any help.

REPLY [9 votes]: Substituting in the definition of $\tau(s)$, that equation is $$\tau(s)=\inf_{\epsilon>0}\tau(s+\epsilon).$$  This means for any $\delta>0$, there exists $\epsilon>0$ such that $\tau(s+\epsilon)<\tau(s)+\delta$.  Since $\tau$ is increasing, this means $\tau(s)\leq\tau(t)<\tau(s)+\delta$ for any $t$ such that $s\leq t<s+\epsilon$.  This exactly verifies the $\epsilon$-$\delta$ definition of right continuity.