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Let $(X,\mathcal{A},\mu)$ a finite measurable space and $f:X\rightarrow[0,\infty]$ a measurable function. I want to show that the integrability of $f$ is equivalent to the convergence of the series

$$\sum_{n=0}^\infty 2^n\mu(\{x\in X\mid f(x)\geq2^n\})$$

I am trying to find a sequence of functions $f_n$ that converge to $f$, so maybe I get the previous series when I compute $$\lim_n\int f_n$$

So far I did not manage to do so, can someone tell me if I am going in the right direction, or point my towards it?

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    Find measurable functions $g,h \colon X \to [0,+\infty]$ such that $g \leqslant f \leqslant h$ and that $\int_X g\,d\mu$, $\int_X h\,d\mu$ have a direct connection to the series. It may help to look at the series $$\sum_{n = 0}^{\infty} 2^n \mu\bigl(\bigl\{ x : 2^n \leqslant f(x) < 2^{n+1}\bigr\}\bigr)$$ instead of the given one. The series are strongly related, one converges if and only if the other does. – 2017-01-21
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    Already took that path but I couldn't make it work. The functions I found were $g=\sum_n 2^n \mathcal{X}_{B_n}$ and $h=\sum_n 2^{n+1} \mathcal{X}_{B_n}$, although I get that the series on the post is equal to $\sum_n(\sum_{m=0}^n2^m)\mu(B_n)$, so I don't know how to relate them. I am calling $B_n$ to the set you defined. – 2017-01-21
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    Wait, $2^n\leq\sum_{m=0}^n2^m\leq2^{n+1}$, so that just finishes the argument right? – 2017-01-21
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    Yes. With the small point that we need to set $h(x) = 1$ on $\{ x : f(x) < 1\}$ to have $f \leqslant h$. But that doesn't impact the argument. – 2017-01-21

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Start by proving that $$ \int_Xf\;d\mu=\int_0^{\infty}\mu(\{x:f(x)\geq t\})\;dt $$ using Tonelli's theorem, and then note that if $2^n\leq t<2^{n+1}$ then $$ \mu(\{x:f(x)\geq 2^n\})\geq \mu(\{x:f(x)\geq t\})\geq \mu(\{x:f(x)\geq 2^{n+1}\})$$