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I'm reading Extreme Value Theory: An Introduction by Laurens de Haan and Ana Ferreira. I've had some trouble following the way they throw around concepts, but this is something I'm really having hard time with.

"Let $f$ be any nondecreasing function and $f^{\leftarrow}$ its left-continuous inverse i.e. $f^\leftarrow(y) = \inf \left\{ x \: \middle| \: f(x) \geq y \right\}$. Check that $\left( f^\leftarrow \right)^\leftarrow = f^-$, with $f^-$ the left-continuous version of $f$."

I've only heard about a concept of a version in the context of stochastic processes.

Matias Heikkilä
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2 Answers2

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If $f$ is a nondecreasing function on $\mathbb R$, then its only discontinuities are jump discontinuities. That is, there is a discrete set $X$ such that $f$ is continuous on $\mathbb R\setminus X$ and for $x\in X$ the onesided limits $\lim_{t\to x^-}f(t)$ and $\lim_{t\to x^+}f(t)$ exist, but differ. For most purposes it does not really matter what $f(x)$ is when $x\in X$. Thus we can replace $f$ with a slightly different function (called a version of $f$). One such version is the leftcontinuous version $f^-$ where $f^-(x)=\lim_{t\to x^-}f(t)$ for all $x\in X$.

Hagen von Eitzen
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Based on a non-decreasing function $f$ you can prescribe function $f^{-}$ by $x\mapsto\sup\left\{ f\left(y\right)\mid y<x\right\} $.

This function is left-continuous and for each $y<x$ it satisfies:$$f(y)\leq f^{-}(x)\leq f(x)$$

Consequently we have $f^{-}(x)=f(x)$ if $f$ is left-continuous at $x$.

drhab
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