Measure theory question Let $f\geq 0$ be a measurable function which is finite almost everywhere.
For each $k\in\mathbb{Z}$, define, $E_k=\lbrace x|f(x)>2^k\rbrace, F_k=\lbrace x|2^k\leq f(x)<2^{k+1}\rbrace$.
Is it true that $\sum_{k=-\infty}^{\infty}2^km(E_k)<\infty$ if and only if $\sum_{k=-\infty}^{\infty}2^km(F_k)<\infty$?

The idea is to bound the two expressions tightly in terms of each other:
$$
\sum_{k=-\infty}^{+\infty} 2^k m(F_k) \leqslant 
\sum_{k=-\infty}^{+\infty} 2^k m(E_k) \leqslant 
2 \cdot\sum_{k=-\infty}^{+\infty} 2^k m(F_k) .
$$
Once this is established, it is obvious that either summation converges if and only if the other does,
The left half of the inequality is obvious since $F_k \subseteq E_k$, so we prove only the right half. 
$$
\begin{align*}
\sum_{k=-\infty}^{+\infty} 2^k m(E_k) 
&\leqslant \sum_{k=-\infty}^{+\infty} 2^k \left( \sum_{j=k}^{\infty} m(F_j) \right)
\\ &= \sum_{j= -\infty}^{\infty} m(F_j) \left( \sum_{k=-\infty}^{j} 2^k \right)
\\ &= \sum_{j= -\infty}^{\infty} 2^j m(F_j) \left( \sum_{k=-\infty}^{0} 2^k \right)
\\ &= \sum_{j= -\infty}^{\infty} 2^j m(F_j) \cdot 2.
\end{align*}
$$
Here the interchange of summations is justified because all terms are nonnegative. 
