Prove the following statement (existence of simple function)

Question

Let $f$ be measurable on $[a,b]$. Then for a given $\varepsilon >0$ and $M>0$ exists a simple function $\phi$ on $[a,b]$ so that $$|f(x)-\phi(x)|<\varepsilon,\; \forall x \in A,\; A=\{x\; : \; |f(x)|>M \}$$

In some proofs they use $f^+$ and $f^-$, but I don't know what they are?

Answer 1

$f^+(x) := \max\{0,f(x)\}$ denotes the positive part and $f^-(x) := \max\{0,-f(x)\}$ the negative part of $f$. The so-defined functions have the property that

• $f^+$ and $f^-$ are measurable (if $f$ is measurable).
• $f^+(x) \geq 0$ und $f^-(x) \geq 0$ for all $x$.
• $f(x) = f^+(x)-f^-(x)$ for all $x$.