Neccesary condition for an integral to be finite Suppose that $(X,\Sigma,\mu)$ is a measurable space and $f$ a non-negative measurable function such that
$$
\int_{X}{f}< +\infty
$$
I want to prove that $\sum_{n=0}^{\infty}{2^n \mu( \left\{{x:f(x) \geq 2^n}\right\}  )} < + \infty$. Let's denote $E_n=\left\{{x:f(x) \geq 2^n}\right\}$, then
$$
\chi_{E_0}+2\chi_{E_1}+\cdots+2^n \chi_{E_n} \leq (n+1)f
$$
Therefore,
$$
\frac{1}{n+1}\sum_{k=0}^{n}{2^k \mu( \left\{{x:f(x) \geq 2^n}\right\}  )}< \int_{X}f
$$
that is, the sequence $\left\{ 2^n \mu(E_n) \right\}_{n \in \mathbb{N}}$ is Cesaro summable so if I could prove that the sequence $\left\{ n2^n \mu(E_n) \right\}_{n \in \mathbb{N}}$ is bounded I'd be done, however I haven't been able to prove it, actually it could be false in which case my idea is wrong.
Can anyone give a hint please?
In advance thank you.
 A: $\sum_{n=1}^{\infty} \mu\left( \{ x \in E : |f(x)| \ge n \} \right) < \infty
$: See Show $f$ : integrable $\Leftrightarrow$ $ \sum_{n=1}^{\infty} \mu\left( \{ x \in E : |f(x)| \ge n \} \right) < \infty $
But $\sum_n <\infty$, $a_n \geq 0$ and $(a_n)$ decreasing implies $\sum_n 2^{n}a_{2^{n}} <\infty$.  See https://en.wikipedia.org/wiki/Cauchy_condensation_test
A: Another way to show this is using Tonelli's theorem.
We have $$\int_{X}|f|\,d\mu= \int_{0}^{\infty}\mu(\{|f|>t\})\,d\lambda(t)$$
The proof goes as follows:-
$$\int_{0}^{\infty}\mu(\{|f|>t\})\,d\lambda(t)=\int_{0}^{\infty}\int_{X}\mathbf{1}_{\{|f|>t\}}\,d\mu(x)\, d\lambda(t)$$
Now use Tonneli's Theorem to say that
$$\int_{X}\int_{0}^{\infty}\mathbf{1}_{\{|f|>t\}}d\lambda(t)\,d\mu(x)=\int_{X}\int_{\Bbb{R}}\mathbf{1}_{\{t\in [0,|f(x)|)\}}\,d\lambda(t)\,d\mu(x)=\\\int_{X}\lambda([0,|f(x)|))\,d\mu(x) =\int_{X}|f(x)|\,d\mu(x) $$
Now this implies that $|f|$ is integrable iff $\int_{0}^{\infty}\mu(\{|f|>t\})\,d\lambda(t)<\infty$ .
But $t\mapsto \mu(\{|f|>t\})$ is a decreasing function and hence due to the integral test  for series . $\displaystyle\int_{0}^{\infty}\mu(\{|f|>t\})\,dt<\infty\iff \sum_{n=1}^{\infty}\mu(\{|f|>n\})<\infty$ . Thus you can now use the Cauchy Condensation test to conclude that $\displaystyle\sum_{n\in\Bbb{N}}2^{n}\mu(\{|f|>2^{n}\})<\infty$
BONUS:- $$\int_{X}|f|^{p}\,d\mu = \int_{0}^{\infty} t^{p-1}\mu(\{|f|>t\})\,d\lambda(t)$$ and the proof is very similar.
