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?

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: $$\sum_{n=1}^{\infty}n\mu(E_n)<\infty\iff\int |f|\;d\mu<\infty$$
