I'm working on a quesiton in Stein's Real Analysis book; it's Chapter 2, Exercise 10.
Suppose that $f \geq 0$ is measurable, in the sense of Lebesgue. Define $E_{k} =\{x \colon 2^k < f(x) \leq 2^{k+1}\}$ and $E_{2^k} = \{x \colon 2^k < f(x)\}$.
I'm having trouble with this implication:
$\sum_{k=-\infty}^{\infty} 2^k m(F_k) < \infty \implies \sum_{k=-\infty}^{\infty} 2^k m(E_{2^k}) < \infty$.
I have the beginning of an idea, which I will post below. If anyone can offer some intuition on how to view these sets or maybe some towards the argument I have that I might be missing. Thanks!
My Attempt: Notice that each $F_k = E_{2^k} \setminus E_{2^{k+1}}$ and so $E_k = F_k \cup E_{2^{k+1}}$, which is a disjoint union. Thus,
$ \sum_{k=-\infty}^{\infty} 2^k m(E_{2^k}) = \sum_{k=-\infty}^{\infty} 2^k m(F_k \cup E_{2^{k+1}}) = \sum_{k=-\infty}^{\infty} 2^k m(F_{k}) + \sum_{k=-\infty}^{\infty} 2^k m(E_{2^{k+1}}) = \sum_{k=-\infty}^{\infty} 2^k m(F_{k}) + \frac{1}{2}\sum_{k=-\infty}^{\infty} 2^{k+1} m(E_{2^{k+1}}), $ ....
If the equalities you have are correct, you have $$ \sum\limits_{k=-\infty}^{\infty} 2^k m(E_{2^k}) = \sum\limits_{k=-\infty}^{\infty} 2^k m(F_k) + \frac{1}{2}\sum\limits_{k=-\infty}^\infty 2^{k+1} m(E_{2^{k+1}}) $$ Notice that the rightmost sum with summand $2^{k+1}m(E_{2^{k+1}})$ is equal to $\frac{1}{2}\sum\limits_{k=-\infty}^{\infty} 2^k m(E_{2^k})$ by a simple change of variables, so you can subtract this term from both sides to get:
$$ \frac{1}{2}\sum\limits_{k=-\infty}^{\infty} 2^k m(E_{2^k}) = \sum\limits_{k=-\infty}^{\infty} 2^k m(F_k) $$
This, of course, assumes that everything is correct to your last equality.
EDIT: This is edited due to the astute point in the comments noticing that if the sum I subtracted was infinite, this operation would not be defined.
Suppose that $\sum\limits_{k=-\infty}^{\infty} 2^k m(2^k) < \infty$. Notice that $$m(E_{2^r}) = \sum\limits_{k=r}^{\infty} m(F_k)$$ so that $$ \sum\limits_{r=-N}^N 2^r m(E_{2^r}) = \sum\limits_{r=-N}^N \sum\limits_{k=r}^\infty 2^rm(F_k) = \sum\limits_{k=-N}^{\infty}\sum\limits_{r=-N}^{\min(k,N)} 2^r m(F_k) = \sum\limits_{k=-N}^{\infty} 2^k m(F_k) \sum\limits_{r=-N}^{\min(k,N)} 2^{r-k} \\ \leq C \sum\limits_{k=-N}^{\infty} 2^k m(F_k) $$ where $C < \infty$ can be chosen independent of $N$; in fact, you could choose $C = \sum\limits_{r = -\infty}^{\infty} 2^{-|r|}$.