Approximation of a Bounded Simple Function by a Continuous Function

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.

Solution

Write

$f = \sum_{j = 1}^m a_j \chi_{E_j \cap [a,b]},$

where $E_j$ is measurable for each $j = 1,\dots,m$, so that we can consider $E_j$. Note that the $E_j$'s are not necessarily disjoint, but who cares? By drawing a simple figure of $E_j \cap [a,b]$ and $U_j$, which is a finite disjoint union of open intervals obtained by applying Littlewood's First Principle, I realised that I should try constructing a simple function $f$ while focusing on $(E_j \cap [a,b]) \triangle U_j$.

In the proof, to deal with the summation sign, we use a union sign.

$f = \psi \triangleq \sum_{j = 1}^m a_j \chi_{U_j} \text{ except on } \bigcup_{j = 1}^m [(E_j \cap [a,b]) \triangle U_j]$

If we’re given that $m \le f \le M$, to find $x \in \bigcup\limits_{j = 1}^m [(E_j \cap [a,b]) \triangle U_j]$ so that $\psi(x) < m$ or $\psi(x) > M$, we need to have overlapping of $E_i \cap [a,b]$ and $E_j \cap [a,b]$, so that $\chi_{(E_i \cap [a,b]) \triangle U_i} + \chi_{(E_j \cap [a,b]) \triangle U_j}$ will give us something interesting.

To see necessity of the introduction of $\bar{\psi}$, consider $f \triangleq \chi_{(0,1)} + \chi_{(1,2)}$ and $\psi \triangleq \chi_{[0,1]} + \chi_{[1,2]}$. Then $f = \psi$ except on $\{0,1,2\}$ and $0 \le f \le 1$, but $\psi(1) = 2 > 1$. One can easily transform a step function into a piecewise linear (i.e. continuous) function $g$.

Remarks

It’s very difficult for a function to be continuous. The result provides a step for proving Lusin’s Theorem, which says that a measurable function can be approximated by a sequence of continuous function. Old news is so exciting! (See this encyclopedia page for explanation.)