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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 each of the following 5 examples, does the Lebesgue integral exist on $(0,\infty)$, and if it does, is it finite?

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

$f(x)=\sin(\frac1{x^2})$

$f(x)=x^{-1}(1-e^{-x})$

$f(x)=\frac{1-e^{-x}}{x}-\frac{1}{1+x}$

$f(x)=x^{-1}(e^{-x}-e^{-1/x})$

I have read a bit about what it means for the integral to exist in general, but I obviously do not fully understand the concept, as when I try to apply it I quickly get confused.

Any help, both in generally understanding the concept, and on any/all of the examples I found would be greatly appreciated.

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    @tomasz: An improper Riemann integral that is not Lebesgue integrable is necessarily not of a nonnegative function. What fails is that the Lebesgue integral is an absolute integral. But you are right that the Riemann integral is really defined for intervals.2012-06-19

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None of these problems is testing your conceptual understanding of Lebesgue integration. All of these functions are continuous, so the Lebesgue integral exists if and only if either $\int_0^\infty f_+(x)\,dx$ or $\int_0^\infty f_-(x)\,dx$ is finite, where $f_+$ is the positive part of $f$ (i.e. $f_+(x) = \max(f(x),0)$), and $f_-$ is the negative part of $f$.

Really these questions are designed to help you practice your calculus skills, particularly the analysis of improper Riemann integrals. In each case, what you want to do is analyze the behavior of the function near $0$ and $\infty$.

For example, in the second problem, the function is bounded as $x\to 0$, and as $x\to\infty$ it's roughly $1/x^2$, so the improper integral converges. (This can be made rigorous using the Limit Comparison Test for improper integrals and L'Hospital's Rule.)

In the third problem, the function is strictly positive, and is roughly $1/x$ as $x\to\infty$, so the improper integral will diverge to $\infty$. Thus, the Lebesgue integral is infinite.

In general, remember that $\displaystyle\int_0^1 \!\!\frac{1}{x^p}\,dx$ converges if and only if $p<1$, and $\displaystyle\int_1^\infty \!\!\frac{1}{x^p}\,dx$ converges if and only if $p>1$. The techniques for checking convergence of improper integrals are the same as those for infinite series (Comparison Test, Limit Comparison Test, etc.)

I would suggest that you play with each of these problems using these sorts of techniques. If you get stuck on any of them, you could post that problem individually as a question.

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    @tomasz Ah, I hadn't noticed that $\pm\infty$ was allowed. I will edit my answer.2012-06-19