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.
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.
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.