# $\int_X |f|^p d\mu < \infty$ $\leftrightarrow$ $\sum_{n=-\infty}^{\infty} a^n \mu$ ({$(x\in X:a^n \leq |f|^p (x) < a^{n+1}$}) < $\infty$

Given a measurable function: $$f:X\rightarrow C$$ on a measured space (X, A, $$\mu$$)

Show that $$\int_X |f|^p d\mu < \infty$$ $$\leftrightarrow$$ $$\sum_{n=-\infty}^{\infty} a^n \mu$$ ({$$(x\in X:a^n \leq |f|^p (x) < a^{n+1}$$}) < $$\infty$$, where $$p\in [1,\infty[$$, $$a\in ]1,\infty[$$, by using disjunct decomposition.

I have absolutely no idea, how to approach this problem. For example for the proof of "$$\rightarrow$$", I don't know, from where I get that given sum..

How do I approach this problem?

HINT For fixed $$a>1$$, $$f\ne0\iff |f|^p>0\iff \exists n\in\Bbb Z$$ such that $$a^n\le |f|^p< a^{n+1}$$.
Ovbserve that $$(0,+\infty)=\bigcup_{n\in\Bbb Z} [a^{n},a^{n+1})$$ and the union is disjoint.
• Where do I consider it exactly? in "$\leftarrow$" or "$\rightarrow$" ? And how do I put them in? Nov 21, 2020 at 16:11
• Both: $\int_X|f|^p=\int_{\{f\ne0\}} |f|^p=\int_{\bigcup A_n}|f|^p=\sum\int_{A_n}|f|^p$. Recall that the union is DISjoiNT. Nov 21, 2020 at 16:13
• I'm sorry, but I don't know how to continue or rather how to put the things, you've given as Hint, into $\sum (\int_{A_{n}} |f|^p) = \sum (\int_{X} |f|^p *1_{A_n})$. I think, approaching $|f|^p$ by the limit of a sequence of step-functions $(f_{k})_{k\in N})$ doesn't help here. Nov 21, 2020 at 16:54
• $a^n\mu(A_n)\le \int_{A_n}|f|^p\le a^{n+1}\mu(A_n)$ Nov 21, 2020 at 16:58