Let it be ($X$,$\Sigma$,$\mu$) a measure space with $\mu(X)<\infty$ and $f:X\rightarrow\mathbb{R}$ a measurable and non negative fuction. Proof that $\int$$f$$d\mu<\infty$ if and only if $\sum \limits_{n=0}^{\infty} \frac{1}{2^n}\mu(\{x\in X:f(x)\geq\frac{1}{2^n}\})$ exists in $\mathbb{R}$.
I've already prove that if $\sum \limits_{n=0}^{\infty} \frac{1}{2^n}\mu(\{x\in X:f(x)\geq\frac{1}{2^n}\})$ exists in $\mathbb{R}$ then $\int$$f$$d\mu<\infty$, but i don't know how to proof the other part. A hint says that i should use that the sequence $(S_n)_{n\in\mathbb{N}}$, with $S_n=\sum \limits_{j=1}^{n2^n} \frac{1}{2^n} \chi_{A_{j,n}}$, where $A_{j,n}=\Bigg\{\begin{array}{lr} f^{-1}[\frac{1}{2^n},\frac{j+1}{2^n}] & \text{if } j\in\{1,...,(n-1)2^n\}\\ f^{-1}[n,\infty] & \text{if } j=n2^n\end{array}$ satisfice that $$\lim_{n \to \infty} S_n =f$$ and that $\forall$$n\in\mathbb{N}:S_n\leq$$S_{n+1}$ , but i don't know hot to use it to proof that if $\sum \limits_{n=0}^{\infty} \frac{1}{2^n}\mu(\{x\in X:f(x)\geq\frac{1}{2^n}\})$ is finite then $\int$$f$$d\mu<\infty$. Can anyone give some hints for the proof? Thanks in advance.