# Riemann integral counterexample to dominated convergence theorem?

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?

• Looks OK. Lebesgue integrals cover a wider class. It is possible that f may be Lebesgue integrable, but not Riemann. – herb steinberg Apr 24 '20 at 21:44

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.
• Nice. And as per my part 2, this example is ruled out if we insist that $f$ be Riemann integrable. Does that added premise make dominated converge hold? – WillG Apr 24 '20 at 23:35