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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?

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Yes.${}{}{}{}{}$ –  Asaf Karagila Jan 31 '13 at 1:42
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Aside: 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 1:46
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@Clayton: Do you mean improper rather than indefinite? –  Jonas Meyer Jan 31 '13 at 1:47
    
@JonasMeyer: That is correct! My mistake... –  Clayton Jan 31 '13 at 2:11
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2 Answers

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.

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Answered in the comments: Yes.

It is a good exercise to try proving it.

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Not true. You need your domain to be bounded. –  Squirtle Nov 1 '13 at 21:01
    
@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 2 '13 at 3:21
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