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

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 $A=\{f > 2^k \}$

share|cite|improve this question
You're right. I just noticed that I left off the $<\infty$ – wrldt Sep 24 '11 at 0:53

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

share|cite|improve this answer
I'm not sure that I follow. I think what you are attempting is outside of what I know in analysis. – wrldt Sep 24 '11 at 9:12
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
It was the Iverson brackets until I looked them up. – wrldt Sep 24 '11 at 18:55

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.