Real Analysis - Lebesgue integrable functions Let $E$ be a measurable set. Suppose $f \geq 0$ and let $E_k=\{x \in E_k|f(x) \in (2^k, 2^{k+1}] \} $ for any integer $k$. If $f$ is finite almost everywhere, then $\bigcup E_k =  \{x \in E |f(x)>0 \},$ and the sets $E_k$ are disjoint.
Prove that $f$ is integrable $\iff$ $\sum_{k \in Z}2^km(E_k)<\infty$ 
and, that the function:
$f(x) =
\begin{cases}
|x|^{-a},  & \text{if $|x| \leq 1$} \\
0, & \text{otherwise}
\end{cases}$
is integrable $\iff a<1$
I've spent a few hours on this problem with some classmates, and I've tried a whole bunch of different methods  of trying to approach it. We're all totally vexed. Since the arrows are biconditional, I've tried proving the statements as written, their inverses, and their converses. Each time I find myself unable to determine the function is integrable because I can't say anything about $m(E_k)$ to determine if the series above converges, even if I split the series in two for positive and negative $k$. I've also thought trying to suppose that the sum is in fact infinite, and seeing if I can show that $f$ is integrable, but that didn't yield any results either.
Can anyone lend me a hand?

We have $$\int_E fdm =\int_{\{x\in E : f(x)>0\}} fdm=\int_{\bigcup_{k\in\mathbb{Z} }E_k } fdm =\sum_{k\in\mathbb{Z}} \int_{E_k} fdm \leq \sum_{k\in\mathbb{Z}} 2^{k+1} m(E_k )\\ =2\cdot\sum_{k\in\mathbb{Z}} 2^{k} m(E_k ).$$
