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?

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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

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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