$f \in L^p \iff$ a certain series converges Let $E_n = \{x \in \mathbb{X} \mid (n-1) \leq |f(x)| < n\}$, where $(\mathbb{X}, \mathcal{A}, \mu)$ is a finite measure space (i.e., $\mu(\mathbb{X}) < \infty$) and $f$ is a measurable function. Prove that $f \in L^p \iff \sum n^p \mu(E_n) < \infty$.
One of the implications is very straightforward to me. It is clear that $\mathbb{X} = \bigcup E_n$, where the union is disjoint. Therefore, we have:
$$\int_\mathbb{X} |f|^p d\mu = \sum_{n=1}^\infty\int_{E_n} |f|^p d\mu$$
By the definition of the sets, we get the inequalities:
$$\sum_{n=1}^\infty \int (n-1)^p \mathbb{1}_{E_n} d\mu \leq \sum_{n=1}^\infty\int_{E_n} |f|^p d\mu < \sum_{n=1}^\infty \int n^p \mathbb{1}_{E_n} d\mu$$
These, in turn, simplify to
$$\sum_{n=1}^\infty (n-1)^p \mu(E_n) \leq \int_{\mathbb{X}} |f|^p d\mu < \sum_{n=1}^\infty  n^p \mu(E_n)$$
Thus, it is immediately apparent that, if the sum converges, then so does the integral, which implies $f \in L^p$.

My problem is in the other implication: $f \in L^p$ implies convergence

I haven't used finite measure yet, so maybe it comes to play. I first thought about expanding $(n-1)^p$, but this would be a mess even if $p$ was an integer (unless in the case where $p = 1$, in which finite measure would imply the result easily).
Does anyone have a hint as per how to proceed?
Thanks in advance!
 A: Hint: $n^p \leqslant 2(n-1)^p$ for large $n$.
A: The actual Lebesgue integral of $|f|^p$ itself is of the form $\sum \xi_n^p \mu(E_n)$ where $n-1 \leq \xi_n<n$. So it is at least $\sum (n-1)^p \mu(E_n)$, so you know that is finite. Then you need to argue that $n^p$ is "not that much bigger" than $(n-1)^p$; obtaining a bound like $n^p \leq c (n-1)^p$ would suffice here.
Note that where finite measure comes into play is when $n=1$.
A: Suppose that $f\in L^{p}$. Note that
\begin{eqnarray*}
 &  & \int|f|^{p}d\mu\\
 & = & \sum_{n=1}^{\infty}\int_{E_{n}}|f|^{p}d\mu\\
 & \geq & \sum_{n=1}^{\infty}\int_{E_{n}}(n-1)^{p}d\mu\\
 & = & \sum_{n=1}^{\infty}(n-1)^{p}\mu(E_{n}).
\end{eqnarray*}
Hence, $\sum_{n=1}^{\infty}(n-1)^{p}\mu(E_{n})<\infty$. Observe that
$\left(\frac{n}{n-1}\right)^{p}\rightarrow1<2$ as $n\rightarrow\infty$,
so there exists $N$ such that $(\frac{n}{n-1})^{p}<2$ whenever $n\geq N$.
It follows that
\begin{eqnarray*}
\sum_{n=N}^{\infty}n^{p}\mu(E_{n}) & \leq & 2\sum_{n=N}^{\infty}(n-1)^{p}\mu(E_{n})\\
 & < & \infty.
\end{eqnarray*}
Since $\mu$ is a finite measure, we also have that $\sum_{n=1}^{N-1}n^{p}\mu(E_{n})<\infty$.
Combining yields $\sum_{n=1}^{\infty}n^{p}\mu(E_{n})<\infty$.
