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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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    change the order of summation, and add with respect to $k$ first.2011-03-29

1 Answers 1

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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.