Bounded measurable function

Question

Suppose $f$ is a bounded measurable function and $m\{x\in \mathbb{R}:f(x)\ne 0 \} > 0$. Prove that for every $\epsilon>0$ there exists a simple function $s$ such that $$\displaystyle \int |f-s|dx<\epsilon.$$ Does any one have idea about how to prove this?

Answer 1

The statement is false.

Let $Q=\mathbb{Q}\cap [0,1]$. Since $Q$ is countably infinite, we can find a bijection $\phi:\mathbb{Z}\to Q$. Define $f:\mathbb{R}\to Q$ to be the function with the value $\phi(z)$ on the half-open interval $(z,z+1]$. It is not hard to show that $f$ is measurable and it is clearly bounded.

Let $s:\mathbb{R}\to\mathbb{R}$ be a simple function with values in the finite set $F$, at least two of which lie in $[0,1]$. Let $$\delta=\min\big\{|y_1-y_2|:y_1,y_2\in F, y_1\neq y_2\big\}.$$

Clearly, $\delta>0$. There are infinitely many elements in $Q$ that differ from each element in $F$ by at least $\delta/3$. Let $P$ be the set of those and define $$B=\bigcup_{p\in P} \bigg(\phi^{-1}(p),\phi^{-1}(p)+1\bigg].$$ Then $$\infty=\infty ~\delta/3\leq\int_B |f-s|dx\leq \int |f-s|dx.$$

Remark: The result would hold under the assumption that $m\{x\in\mathbb{R}:f(x)\neq 0\}<\infty$.