Using the Monotone Convergence Theorem Let $(X, \mathbb{F}, \mu)$ be a finite measure space. If $f$ is measurable, let $E_n = \{x \in X: (n-1) \leq |f(x)| < n\}$. Show that $f$ is integrable if and only if $\sum_{n=1}^{\infty} n\mu(E_n) < \infty$.
I proved the above using the monotonicity of the integral. However, is there a way to prove this using the monotone convergence theorem? I feel like you should be able but I do not see it.
EDIT
I just realized I did not use the fact that the space has finite measure. How does that fact play a role?
 A: Since $f= f^+-f_-$ we can assume that $f$ is nonegative.
Assume that $\sum_n n\mu(E_n) <\infty$. Let $ f_n = \sum_{i=1}^n f(x) \chi_{E_i}(x)$. Then $f_n \le f_{n+1}$ and $f_n \to f$ and each $f_n$ is measurable.  Due to the monotone convergence thm $\lim_n \int f_n d\mu \to \int f d\mu$. It is clear that $f\le \sum_{i=1}^\infty i \chi_{E_i}(x):=g(x) $ and so $$\int fd\mu\le \int g d\mu = \sum_{i=1}^\infty i \mu(E_i)<\infty$$
due to definition of the integral. ($g$ is the limit of an increasing sequence of simple functions). 
Now assume that $f$ is integrable. Then, due to the finiteness of the measure $f+1$ is also integrable. But $g\le f +1$, hence $$ \sum_{n} n\mu(E_n) = \int gd\mu \le \int (f +1)d\mu < \infty.$$
A: Observe that: $$\sum_{n=1}^{\infty}(n-1)1_{E_n}\leq |f|\leq \sum_{n=1}^{\infty}n1_{E_n}$$Taking the integral on both sides leads to:$$\sum_{n=1}^{\infty}n\mu(E_n)-\sum_{n=1}^{\infty}\mu(E_n)=\sum_{n=1}^{\infty}(n-1)\mu(E_n)\leq\int |f|\;d\mu\leq \sum_{n=1}^{\infty}n\mu(E_n)$$where $\sum_{n=1}^{\infty}\mu(E_n)=\mu(X)<\infty$.
This shows that: $$\sum_{n=1}^{\infty}n\mu(E_n)<\infty\iff\int |f|\;d\mu<\infty$$
