# Condition for $\int f < \infty$ where $f$ is nonnegative and measurable

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 $<\infty$ –  wrldt Sep 24 '11 at 0:53

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 $2^{i-1}<f(x)\le 2^{i}$ for a given integer $i$ and compute $g(x)=\sum\limits_k\,2^k\cdot[k\le i-1]=2^i$.
You ARE following a course in measure theory, aren't you? So... you might care to explain, either what step causes you trouble, or what you know in analysis, or both. Note that the solution above uses basic properties of measures like $\int\alpha\mathbf 1_A=\alpha m(A)$ and $\int (f_1+f_2)=\int f_1+\int f_2$ and the fact that $\sum_{k\ge0}2^{-k}=2$, and nearly nothing else. Oh, and we used Iverson bracket. –  Did Sep 24 '11 at 12:21