Problem
Today, I was puzzled by a remark of a proof that a simple function $f:[a,b] \to \R$ is equal to a step function $\psi:[a,b] \to \R$ (resp. a continuous function $g:[a,b] \to \R$) except on a set of arbitrarily small measure $\epsilon$.
If $m \le f(x) \le M \;\forall\, x\in [a,b]$, then $\bar{\psi} \triangleq (m \vee \psi) \wedge M$ and $\bar{g} \triangleq (m \vee g) \wedge M$ are step function and continuous function which satisfy $m \le \bar{\psi}$ and $m \le \bar{g} \le M$ so that $f = \bar{\psi}$ and $f = \bar{g}$ except on a set of measure less than $\epsilon$ respectively.
I can verify the above remark, but I wonder why we need to define $\bar{\psi}$ and $\bar{g}$ so as to guarantee that they are bounded below and above by $m$ and $M$ respectively.