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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?

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





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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Can you answer any of these questions? Where are you stuck? – tomasz Jun 18 '12 at 22:30
@tomasz, I am confused on the rigorous definition of what it means for the integral to exist. – Justin Jun 18 '12 at 22:34
The integral of a nonnegative continuous function exists always and coincides with the Riemann integral. The function is integrable, if this integral is finite. For general continuous functions, the integral exists if the positive and negative parts do not both have an infinite integral. If both parts are finite, the function is integrable. – Michael Greinecker Jun 18 '12 at 22:41
@MichaelGreinecker: what you're saying is almost correct, except only on finite intervals, as Riemann integral is not defined otherwise. There is improper Riemann integral, but there are cases where it exists, and the Lebesgue integral doesn't, just as there are cases where the Lebesgue integral exists, and Riemann (proper or not) doesn't. – tomasz Jun 19 '12 at 1:11
@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. – Michael Greinecker Jun 19 '12 at 6:51
up vote 2 down vote accepted

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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thank you. I guess I was trying to find something that wasnt there. – Justin Jun 19 '12 at 0:13
The question admits $\pm$infinity as valid integral, so what you said is not exactly true. Lebesgue integral might still exist even if the improper integral does not converge absolutely. For example, in the problem 3. you analyzed the integral exists (the function is positive!), even though the function is not integrable. For continuous functions, the integral exists if one of the parts' -- positive and negative -- improper integrals converges to a finite limit, and is integrable if both do. – tomasz Jun 19 '12 at 1:18
@tomasz Ah, I hadn't noticed that $\pm\infty$ was allowed. I will edit my answer. – Jim Belk Jun 19 '12 at 16:06

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