# 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 $\sum_{k=-\infty}^{\infty}2^km(F_k)<\infty$?

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You say this is for homework, how can we help? What do you get about the question, what are you stuck with? – David Kohler Mar 29 '11 at 18:12
hint: $\sum_{l \geq k+1}m(F_l) \leq m(E_k) \leq \sum_{l \geq k}m(F_l)$. – Zarrax Mar 29 '11 at 18:12
I'm stuck with how can I control $\sum_{k=-\infty}^\infty \sum_{l\geq k}2^k m(F_k)$. Becasue there are too many repeated terms. – user8484 Mar 29 '11 at 18:41
We have $2^km(E_k)\leq \sum_{l\geq k}2^km(F_l)\leq \sum_{l\geq k} 2^lm(F_l)$. If we denote $s_k=\sum_{l\geq k} 2^lm(F_l)$, then we know $s_k$ converges to $0$ but we do not know how fast $s_k$ converges to 0 which I need to conclude that $\sum_{k=-\infty}^\infty s_k<\infty$. – user8484 Mar 29 '11 at 18:59
change the order of summation, and add with respect to $k$ first. – Zarrax Mar 29 '11 at 21:58

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.