# Show that $f$ is measurable if $f\lvert f \rvert^{p-1}$ is measurable

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.

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?
• Right! If I understand correctly the restriction $f'$ of $f$ to $\Omega'=\{ x : f(x) \neq 0 \}$ is $\Omega' \cap \mathscr{B}$ measurable. Thus the preimage of any borel set under $f$ must be either $\Omega' \cap B$ or $(\Omega' \cap B) \cup (\Omega \setminus \Omega' )$ for some $B \in \mathscr{B}$, in either case an element of $\mathcal{F}$. May 31, 2020 at 9:05