Is every Riemann-integrable function lebesgue integrable? I thought every Riemann integrable function is lebesgue integrable. But here, a user comments that the statement is not correct.  I'm confused.  Can someone clarify?
 A: Define $\inf $ to be the infimum over all upper sums of a function $ f $ and $\sup $ to be the supremum over all lower sums. Every step function is also a simple function, every upper sum is a simple function that is bigger than $f$, and every lower function is a simple function less than $f$, giving
$$\sup \leq\sup_{\text {Lesbegue integrable}}\leq \inf_{\text {Lesbegue integrable}}\leq \inf$$
If $f$ is Riemann integrable, the first and last quantities agree, so $f$ must be Lebesgue integrable as well with the same value for the integral. Q.E.D.
A: No. This does not have to always hold on unbounded intervals. 
$$\int_{0}^{\infty} \frac{\sin{x}}{x} \, dx$$ exists in the sense of Riemann, but does not exist in the sense of Lebesgue as 
$$\int_{(0,\infty)} \frac{\sin{x}}{x} \, d\mu. $$
This has to do with the fact that there is no "conditional convergence" in Lebesgue Integration.
A: A characterization of Riemann integrable function is that $f$ is Riemann integrable iff it is continuous almost everywhere.
Now, since in Riemann integration we always talk about integration on bounded intervals, and in Lebesgue integration we do not differentiate between functions that are equal almost everywhere, and since continuous functions are Lebesgue integrable on bounded intervals, we have our result.
