Lebesgue integrable function characterization If $f\in M^{+}(X,S)$ is bounded, prove that
$$ \int f d\mu <\infty\ \iff \sum_{n=o}^{\infty} \frac1{2^n} \mu (\{x\in X: f(x)\ge \frac1{2^n}\})< \infty$$
Obs: After starting any part of the proof I realized that, since $f\in M^{+}(X,S)$ and bounded, it exist an $M\in \Bbb R$ such that $0\le f(x)< M\ \forall x\in X$, so integrating we get that $0\le \int f d\mu<M\mu(X)$. Then $\int f d\mu <\infty$ follows without assuming $\sum_{n=o}^{\infty} \frac1{2^n} \mu (\{x\in X: f(x)\ge \frac1{2^n}\})< \infty$, if $\mu(X)<\infty$. 
So, thinking that $\mu(X)=\infty$ is possible I started with the second part ($\Leftarrow$|):

Provided that $\sum_{n=o}^{\infty} \frac1{2^n} \mu (\{x\in X: f(x)\ge \frac1{2^n}\})< \infty\\$,
it follows that $\mu(E_n)<\infty\ \forall n\in\Bbb N$, where $E_n=\{x\in X: f(x)\ge \frac1{2^n}\}\ \forall n\in \Bbb N$

From here I got stuck . The only thing I have done is that the limit of $E_n$ is $E=\{x\in X: f(x)>0\}$, since $(E_n)$ is an increasing sequence and, thus, by taking the union we get $E$. I can't manage to see I way to connect $\sum_{n=o}^{\infty} \frac1{2^n} \mu (E_n)< \infty$ with some kind of convergence to $f$.
Only that $\int \sum_{n=o}^{\infty} \frac1{2^n} \Bbb 1_{E_n} d\mu=\sum_{n=o}^{\infty} \frac1{2^n} \mu (E_n)< \infty$
 A: Proof of $\Leftarrow$:
Note that, since $f$ is positive, $X-E=\{x\in X:f(x)=0\}$. 
For $n\geq 1$, let $F_n:=E_{n}-E_{n-1}$. Then for all $n\geq 1$, $F_n$ are measurable and $F_n=\left\{x\in X:\frac{1}{2^{n}}\leq f(x)< \frac{1}{2^{n-1}}\right\}$, so $\mu(F_n)<\mu(E_n)$. We can show
\begin{equation}
E  = \left(\cup_{n\geq 1}F_n\right)\cup E_0
\end{equation}
Using this, we have
\begin{eqnarray}
\int_Xfd\mu & = & \int_{X-E}fd\mu+\int_Efd\mu\\
& = & \int_{\{f=0\}}fd\mu+\int_E fd\mu\\
& = & \int_E fd\mu
\end{eqnarray}
Using the sets $F_n$ are disjoint, by the monotone convergence theorem we have
\begin{eqnarray}
\int_Xfd\mu & =  & \int_E fd\mu\\
& = & \int_{\left(\cup_{n\geq 1} F_n\right)\cup E_0}fd\mu\\
& = & \sum_{n\geq 1} \int_{F_n} fd\mu +\int_{E_0} fd\mu
\end{eqnarray}
You already showed the set $E_0$ has finite measure. Adding that $f$ is bounded by $M$, 
\begin{equation}
\int_{E_0}fd\mu<M\mu(E_0)<\infty
\end{equation}
Also, for any $n\geq 1$, if $x\in F_n$ then $f(x)<\frac{1}{2^{n-1}}=\frac{2}{2^n}$, so
\begin{equation}
\int_{F_n}fd\mu<\frac{2}{2^n}\mu(F_n)<\frac{2}{2^n}\mu(E_n)\Rightarrow \sum_{n\geq 1} \int_{F_n} fd\mu<2\sum_{n\geq 1} \frac{\mu(E_n)}{2^n}<\infty
\end{equation}
Then 
\begin{equation}
\int fd\mu<2\sum_{n\geq 1} \frac{\mu(E_n)}{2^n}+M\mu(E_0)<\infty
\end{equation}
