Riemann integral counterexample to dominated convergence theorem?

Question

The dominated convergence theorem (and similar theorems) is often claimed to be what makes Lebesgue integration superior to Riemann integration. But we also have the result that any (positive) Riemann-integrable function is Lebesgue integrable and that both integrals agree in this case. Thus it seems naively reasonable that the dominated convergence theorem should apply to Riemann integrals too.

Let $f_n:I\to\mathbb R^+$, where $I$ is any interval in $\mathbb R$ (possibly infinite), be a sequence of Riemann-integrable functions that converge pointwise to $f$. Suppose further that $|f(x)|\leqslant |g(x)|$ on $I$ for some Riemann-integrable function $g(x)$.

1. Do we have that $\lim_{n\to\infty}\int_{I,\text{Riemann}}f_n(x)dx=\int_{I,\text{Riemann}}f(x)dx$ ?
2. What if we add the premise that $f$ is also Riemann integrable?

If the answer to (1) turns out to be "no," what's a counterexample?

Answer 1

For example, let $r_n$ be an enumeration of the rationals in $[0,1]$, and let $f_n$ be the indicator function of $\{r_1, \ldots, r_n\}$. Then $f_n$ are Riemann integrable with $|f_n| \le 1$, but $f_n$ converges pointwise to the indicator function of the rationals, which is not Riemann integrable.

Answer 2

The first part has been seen to be false.

For the second part, it seems to be false, too. Let $I=[0,1]$ and $f_n(x)=n\, \mathbb I([0,\frac1n])$. Then $f_n$ converges pointwise to $f=0$ which is integrable. However $\lim\int f_n=1$ while $\int f=0$. I think the proper assumption should be $f_n\leqslant g$ instead of $f\leqslant g$. Then we can apply dominated convergence theorem and conclude by the fact that Riemann integral and Lebesgue integral conincide. (In fact the proof would be tough if you don't use any Lebesgue theory.)