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This is a practice problem from Carothers p. 321.

Let $f$ be nonnegative and measurable. Prove that $\int f < \infty$ if and only if $\sum_{-\infty}^\infty 2^km(\{f > 2^k\}) < \infty .$

One thing I noticed right away was that $\int 2^k \chi_A = 2^km(\{f > 2^k\})$ where $A=\{f > 2^k \}$

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    You're right. I just noticed that I left off the <\infty2011-09-24

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Oh, the power of slices...

Consider $g(x)=\sum\limits_k\,2^k\cdot[f(x)>2^k]$. Then $f\le g\le 2f$ and $\sum\limits_k\,2^km(f>2^k)$ is the integral of $g$. You are done.

To prove that $f(x)\le g(x)<2f(x)$ when $f(x)\ne0$, assume that $2^{i-1} for a given integer $i$ and compute $g(x)=\sum\limits_k\,2^k\cdot[k\le i-1]=2^i$.

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    It was the Iverson brackets until I looked them up.2011-09-24