If $f$ is non negative measurable, $ \int f \lt \infty $ if and only if $\sum_{n=-\infty}^{\infty} 2^{n} m{(f\gt 2^n)} \lt\infty $

Question

Let $f$ be non negative measurable. Prove that $ \int f \lt \infty $ if and only if $\sum_{n=-\infty}^{\infty} 2^{n} m{(f\gt 2^n)} \lt\infty $.

This is a very popular question in Lebesgue integration. I have seen similar type of question so frequently in which sum varies from $0$ to $\infty$. But I do not have valid ideas to prove the current above one.

Answer 1

The sum $s = \sum_{2^n<a} 2^n$ evaluates to $2^n$ where $n$ is the smallest integer such that $2^n\ge a$. Hence, $s/2\le a\le 2s$.

The above observation made, the proof consists of introducing $$g=\sum_{n}2^n \chi_{\{f>2^n\}}$$ and using the inequalities $$g/2 \le f \le 2g $$