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Is it true that the infimum of all piecewise constant functions that majorize a function $f$ is $f$ itself? If yes, how do I show it formally?

Definition 1. Let $I$ be bounded interval. Let $f:I \to \mathbb{R}$ be bounded function. We call $f$ piecewise constant on $I$ iff there exists some partition $P$ of $I$, such that $f$ is constant on each $J\in P$.

Definition 2. .... We say, f majorizes g, iff $\forall x \in I: f(x) \geq g(x)$.

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You have not put any condition on $f$ except boundedness. If $f$ is assumed to be continuous then the result is true. In general it is false. For example, if $f(0)=-1$ and $f(x)=0$ for $x \neq 0$ then, for any pieceiwise constant function $g$ with $g \geq f$ we have $g(0) \geq 0$ so the infimum of all such functions is not equal to $f$.

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  • $\begingroup$ Thank you very much, I get it now! How about when $f$ is Riemann integrable? $\endgroup$ – Cebiş Mellim May 2 '18 at 9:36
  • $\begingroup$ The function $f$ in my example is Riemann integrable, right? $\endgroup$ – Kavi Rama Murthy May 3 '18 at 10:46

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