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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?

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

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  • $\begingroup$ 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? $\endgroup$ – WillG Apr 24 '20 at 23:35

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