# Convergence of Riemann integrals that do not hold for Lebesgue integrals

I am interested in convergence results that are true for Riemann-integrable functions but fail for Lebesgue-integrable functions. I know three of these, which happen to be closely related.

• Convergence of Riemann sums of a Riemann-integrable function $f : [a,b] \rightarrow {\bf R}$.

• Convergence of means of Riemann-integrable functions along equidistributed sequences (Weyl criterion).

• The weak convergence of $P_n$ to $P$ implies $\int f dP_n \rightarrow \int f dP$ for all $P$-Riemann-integrable $f$ ($P$-Riemann integrability meaning that the set of discontinuities of $f$ is negligible wrt $P$).

Any others?

As I learned and discussed here, if $f\colon\mathbb{R}\to\mathbb{R}$ is Riemann-integrable, and if $\mathscr{M}_n(f) := \frac{1}{n}\sum_{k=0}^{n-1}\; f\big(\!x+\frac{k}{n}\big)$ denotes the average of the translates of $f$ by multiples of $1/n$, then $\mathscr{M}_n(f)$ converges uniformly to the average value $\int_0^1 f$ of $f$ (a constant function), but convergence a.e. does not hold for $f$ Lebesgue-integrable, even if it is assumed to be bounded (Walter Rudin, "An Arithmetic Property of Riemann Sums", Proc. Amer. Math. Soc. 15 (1964), 321–324).
• @coudy It is of course a kind of convergence of Riemann sums, but note that for $f$ R-integrable we get uniform convergence, so it's a bit stronger in that sense; and, of course, the negative result for $f$ L-integrable is also not subsumed by general nonconvergence because this is a very particular kind of Riemann sums. – Gro-Tsen Jun 29 '16 at 20:03