Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.