Does Riemann integrable imply Lebesgue integrable?

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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Yes.${}{}{}{}{}$ – Asaf Karagila Jan 31 '13 at 1:42
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
@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
So both of the integral approaches are equivalent? – user117913 Nov 3 '14 at 15:51

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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