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I came across the following interesting problem in my self-study:

Let $f: X \rightarrow [0, \infty]$ be a measurable function. Assume that $\mu(X) < \infty$. Prove that $\int f \; d\mu < \infty$ if and only if $$\sum_{n=1}^\infty 2^n \mu(x \in X : f(x) \geq 2^n) < \infty.$$

I am having trouble thinking of where to begin in proving this result, and wanted to see if anyone visiting had some suggestions on how to proceed.

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I'd have written $\{x\in X: f(x) \ge 2^n\}$ rather than $(x\in X: f(x) \ge 2^n)$. Using curly braces in that way is standard. Are you sure it wasn't written that way? –  Michael Hardy Feb 9 '12 at 16:55
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3 Answers

up vote 5 down vote accepted

This is a consequence of the following two facts, which you might wish to prove carefully:

  1. The series is the integral of the function $g=\sum\limits_{n\geqslant1}2^n\mathbf 1_{A_n}$ with $A_n=\{f\geqslant 2^n\}$.
  2. One has $\frac12g\leqslant f\leqslant2+g$ and the function $2$ is integrable.
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I really like this approach. I have deduced 1, but 2 is still eluding me. How should that argument proceed? –  Vulcan Feb 10 '12 at 17:00
Try to prove the double inequality $\frac12g\leqslant f\leqslant 2+g$, separately on $\{f\lt2\}$ and on $\{2^n\leqslant f\lt2^{n+1}\}$ for each $n\geqslant1$. –  Did Feb 10 '12 at 17:12
Thank you very much, your suggestion helped me to think about the proof of 2 in the right way! –  Vulcan Feb 10 '12 at 23:10
Great. Well done. –  Did Feb 11 '12 at 2:24
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Warning: The following may contain errors since I myself is learning measure theory at the moment. I have just mentioned what I think are the basic ideas bellow. Thus if you wish you can fill in the details your self.

For one of the directions you can use:

We see that if $∫fdμ<∞$ holds then by markovs inequality we have $2^nμ(x∈X:f(x)≥2^n)\leq ∫fdμ<∞$ for every $n$. Hence $μ(x∈X:f(x)≥2^n)\rightarrow 0$ as $n\rightarrow 0$. Now applying abels criterium of convergence for series it follows that the series converges.

For the reverse note that if the series converges then $μ(x∈X:f(x)≥2^n)\rightarrow 0$ as $n\rightarrow 0$. Now define $A_n:=\{x \in X | f(x)\geq2^n\}$ use downward monotone convergence of measures on these sets to conclude that the set $\{f=\infty\}$ has measure zero. Now since $\mu(X/\cap _nA_n) =\mu(X)<\infty$ and $f(x)$ is finite on $X/\cap _nA_n$ the claim follows.

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It makes much more sense if you draw a topographic picture. If we define $A_n = \{x\in X\ |\ f(x)\geq 2^n\}$ and $A_0 = X$, then we can write X as a disjoint union $\displaystyle X = \bigsqcup_{n=1}^{\infty} (A_{n-1}\setminus A_n)$. Consequently, $$ \int_X f d\mu = \sum_{n=1}^{\infty} \int_{A_{n-1}\setminus A_n} f d\mu $$

For the "only if" part, observe that $$ \displaystyle \int_X f d\mu \geq \sum_{n=1}^{\infty} \int_{A_{n}\setminus A_{n+1}} f d\mu \geq \sum_{n=1}^{\infty} 2^n(\mu(A_n) - \mu(A_{n+1})) = \mu(A_1) + \sum_{n=1}^{\infty} 2^{n-1} \mu(A_n) $$

For the "if" part, observe that $$ \displaystyle \int_X f d\mu = \sum_{n=1}^{\infty} \int_{A_{n-1}\setminus A_n} f d\mu \leq \sum_{n=1}^{\infty} 2^n \mu(A_{n-1}) = 2\mu(X) + 2\sum_{n=1}^{\infty} 2^n \mu(A_n) $$

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