Finite lebesgue Integral Hi guys I've been trying to prove this for a very long time, if someone could help me i would appreciated very much! let $(X,S,\mu)$ be a mesurable space, if $\mu(X)$ is finite and $f$ is a mesureble  non negative function then:
if $\int f d\mu$ is finite then $\sum_{n=0}^{\infty} 2^n \mu(\{ x \in X \vert f(x) \ge 2^n \})$ is finite 
 A: Define the family $(a_{n,k},n,k\geqslant 0)$ of non-negative real numbers by  $$a_{n,k}:=2^n\mu\{x\in X,2^k\leqslant f(x)\lt 2^{k+1}\}[k\geqslant n].$$
Then 
$$\sum_{n\geqslant 0} 2^n\mu\{x\in X,|f(x)|\geqslant 2^n\}=\sum_{n,k\geqslant 0}a_{n,k}=\sum_{k=0}^\infty\sum_{n=0}^\infty a_{n,k}=\sum_{k=0}^\infty2^{k+1}\mu\{x\in X,2^k\leqslant f(x)\lt 2^{k+1}\}\leqslant 2\int_X|f|\mathrm d\mu.$$
A: You can make use of Cauchy condensation test.
A: Here is my attempt:
$$\because \int fd\mu<\infty$$
$$\mu\left(\{f=\infty\right\})=0$$ 
$$Let\ \epsilon>0\ be\ given.$$
$$Define\ A_n=\left\{x\in X|f(x)\geq2^n\right\}$$
$$Note\ that\ A_0 \supseteq A_1 \supseteq A_2 \ldots$$
$$\mu(A_0)<\infty$$
$$\cap_{n=0}^{\infty}A_n=\left\{f=\infty\right\}$$
$$\therefore lim_{n \to \infty}\mu(A_n)=\mu(\cap_{n=0}^{\infty}A_n)=\mu(\left\{f=\infty\right\})=0$$
$$\therefore \exists\ N\in \mathbb{N}\ s.t\ \forall\ n>N,\ \mu(A_n)<{\epsilon \over 2^{2n}}$$ 
$$\sum_{n=0}^{\infty}2^n\mu(\left\{x\in X|f(x)\geq2^n\right\})$$
$$=\sum_{n=0}^{N}2^n\mu(\left\{x\in X|f(x)\geq2^n\right\})+\sum_{n=N+1}^{\infty}2^n\mu(\left\{x\in X|f(x)\geq2^n\right\})$$
$$\leq\sum_{n=0}^{N}2^n\mu(\left\{x\in X|f(x)\geq2^n\right\})+\sum_{n=N+1}^{\infty}2^n{\epsilon \over 2^{2n}}$$
$$\leq\sum_{n=0}^{N}2^n\mu(\left\{x\in X|f(x)\geq2^n\right\})+\sum_{n=0}^{\infty}{\epsilon \over 2^{n}}$$
$$=\sum_{n=0}^{N}2^n\mu(\left\{x\in X|f(x)\geq2^n\right\})+2\epsilon$$
The resulting sum is bounded.
Therefore the result follows.
Please comment on my attempt as I am also learning Measure Theory recently.:)
