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Problem:

At the request of another user, I am taking an older question and specifically addressing one problem.

I am self-learning about Lebesgue integration, and am just starting to try and apply some examples of the existence of the integral.

For the function below, does the Lebesgue integral exist on $(0,\infty)$, and if it does, is it finite?

$f(x)=\sum_{k=0}^\infty e^{-2^kx}$

Since this is self-learning from scratch, I would be grateful if someone could help me break this down bit by bit:

1- What does it mean for the integral to "exist"? Is this just saying that $\int f(x)$ is finite?

2- How do you calculate the integral explicitly?

Any help is always appreciated. Thanks!

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A simple application of dominated convergence theorem shows that, if $\sum \int |f_n(x)| \; dx$ is finite, then $\sum f_n(x)$ converges a.e., it is integrable, and its integral can be calculated termwise. This applies to your function. –  sos440 Jun 22 '12 at 5:11
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2 Answers 2

up vote 2 down vote accepted
  1. The integral exists means it is finite!
  2. Set $S_n(x)=\sum_{k=0}^n\exp(-2^kx)$. Then $S_n (x)\le f(x)$ and $S_n(x) \to f(x)$ for every $x>0$. Thus \begin{eqnarray} \int_0^\infty f &=&\lim_{n\to \infty}\int_0^\infty S_n=\lim_{n\to \infty}\sum_{k=0}^n\int_0^\infty\exp(-2^kx)dx=\lim_{n\to\infty}\sum_{k=0}^n\frac{1}{2^k}\cr &=&\lim_{n\to\infty}\frac{1-(1/2)^{n+1}}{1-1/2}=\frac{1}{1-1/2}=2 \end{eqnarray}
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Thank you for the help.If I understand correctly the first equality is the step we can make based on Monotone Convergence Theorem, right? If that is the case, $S_n$ must meet 3 criteria: it must be non-negative, measurable, and monotone increasing to $f$. Could you help me (or show) how to rigorously prove measurability and monotone-increasing? Also, please correct me if I have the wrong understanding. –  Justin Jun 23 '12 at 20:37
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(i) Clearly $S_n(x)>0$ for every $x$ (it is the sum of expoents!). (ii) $S_{n+1}(x)=S_n(x)+\exp(-2^{n+1}x)>S_n(x)$, i.e. for every $x>0, \ S_n(x)$ is increasing. (iii) Clearly $S_n(x) \to f(x)$ for every $x>0$. For each $n$ the function $S_n$ is continuous, therefore it is measurable. –  Mercy Jun 23 '12 at 20:57
    
Cheers! and thank you! This is all very new to me and self-taught, so I am just glad I apparently identified the application of MCT in the first place :) –  Justin Jun 23 '12 at 20:58
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  1. Let $f_+$ be the positive part of $f$ and $f_-$ it's negative part, so that $f = f_+ + f_-$. The integral exists if $f$ is measurable and at most one of the integrals of $f_+$ and $f_-$ are infinite.

  2. Can you solve the integral for a fixed $k$ disregarding the sum? When can you change places between integral and sum?

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$f>0$ therefore $f=f_+, \ f_-=0$. –  Mercy Jun 22 '12 at 9:44
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