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Suppose we have a right-continuous function $f$ defined on $[0,\infty)$. I now would like to prove if $\lambda >0$ and $t\geq 0$ we have that \begin{equation} \left\{ \inf \{ s\geq 0 : f(s) \geq \lambda \} \leq t \right\} = \left\{ \sup_{s\leq t} f(s) \geq \lambda \right\}. \end{equation} Is this in general true and if so how could I prove this?

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In general, this isn't true. For example, take $t:=2$, $\lambda:=1$, and $$ f(x):=\left\{\begin{array}{cc}x, & 0\le x<1,\\ 0, & 1\le x<3,\\ 1, & x\ge 3.\end{array}\right. $$ Then $$ \sup_{s\le t} f(s)=\sup_{s\le 2} f(s)=1\ge \lambda $$ but $$\inf \{ s\ge 0: f(s)\ge \lambda\}=\inf \{s\ge 0: f(s)\ge 1\}=\inf \{s: s \ge 3\}=3\not\le t.$$

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