Suppose a definite integral exists in the Riemann sense. Does that mean the integral exists as a Lebesgue integral, and do we get the same result either way? ------- BTW: I have a MS in Electrical Engineering and a strong interest in math. I had one semester of real analysis 25 years ago, I tried to learn Lebesgue integration on my own by reading a book on real analysis, and that was a few years ago. Hence, I don't have a solid grasp of the subject.
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2Yes.${}{}{}{}{}$ – Asaf Karagila Jan 31 '13 at 01:42
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4Aside: It is not true for indefinite integrals. In particular, one can consider $$\int_0^\infty\sin(x^2)\space dx.$$ – Clayton Jan 31 '13 at 01:46
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4@Clayton: Do you mean improper rather than indefinite? – Jonas Meyer Jan 31 '13 at 01:47
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@JonasMeyer: That is correct! My mistake... – Clayton Jan 31 '13 at 02:11
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So both of the integral approaches are equivalent? – user117913 Nov 03 '14 at 15:51
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This is true for "properly" Riemann integrable functions $f: [a,b] \rightarrow \mathbb{R}$, a fact which is established in all standard treatments of the Lebesgue integral.
However, there are improperly Riemann integrable functions $f: [0,\infty) \rightarrow \mathbb{R}$ which are not Lebesgue integrable. The most standard counterexample has already been discussed on this site: see here.
Pete L. Clark
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Answered in the comments: Yes.
If it is improperly Riemann integrable, that means it is rather a limit of Riemann integrals, and it is a limit of Lebesgue integrals in the same manner. The difference is that for Riemann integrals, that is the only way to define an integral in the "improper" case, whereas for Lebesgue integration there is a definition that in general might converge without having to restrict the domain and then take the limit. Pete's answer points out why the distinction needs to be made.
Jonas Meyer
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1@Squirtle: You are referring to improper Riemann integrals, not ordinary Riemann integrals. An improper Riemann integral on an unbounded domain is defined as a limit of Riemann integrals on bounded domains. What you point out has already been covered twice in this thread, but thank you for pointing out the lack of clarity in my answer. This distinction is also referred to here. – Jonas Meyer Nov 02 '13 at 03:21