This is exercise 2 from chapter 15 in Bauer's book Measure and Integration Theory:
For an arbitrary measure space $(\Omega,\mathcal{F},\mu)$ and $1\leq p<\infty$ show that a real function $f$ on $\Omega$ is $p$ integrable if and only if $f\lvert f \rvert^{p-1}$ is integrable. (In the "if" direction measurability of $f$ itself is not part of the hypothesis.)
Since $\lvert f \rvert^{p}= \bigg\lvert f\lvert f \rvert^{p-1} \bigg\rvert $ the assertion clearly holds if $f$ is measurable. But in "if" direction how I show that measurability of $f$ follows from the measurability of $f\lvert f \rvert^{p-1}$?
Any help is greatly appreciated.
Best Answer
Hints: $|f|f|^{p-1}|=|f|^{p}$ is measurable. Composing with the continuous function $x \to x^{1/p}$ we see that $|f|$ is measurable. In particular $\{x: f(x)=0\}=\{x: |f(x)|=0\}$ is measurable. On the set $\{x: f(x)\neq 0\}$ we can write $f=\frac 1{|f|^{p-1}} f|f|^{p-1}$. Can you finish?