The following question is based on discussion involving Dominated Convergence Theorem presented by Ross.
Let $f(x) = 1$ for rational $x$ in $[a, b]$ and $f(x) = 0$ for irrational $x$ in $[a, b]$. Let $(x_{k})_{k \in \mathbb{N}}$ be an enumeration of the rationals in $[a, b]$, and define $f_{n}(x_{k}) = 1$ for $1 \leq k \leq n$ and $f_{n}(x) = 0$ for all other $x \in [a, b]$.
- Then $f$ is not integrable because the upper and lower Darboux integrals do not agree.
- However, $f_{n}$ is integrable. Let $g(x) = 0$ for $x \in [a, b]$. Then $f_{n}$ is integrable because $f_{n}(x) = g(x)$ except for finitely many $x \in [a, b]$.
- Additionally, $\lim_{n \rightarrow \infty}f_{n} = f$
Then by Dominated Convergence Theorem (Theorem 33.11 in Ross), if there exists $M > 0$ such that $\lvert f_{n}(x) \lvert \leq M$ for all $n$ and all $x \in [a, b]$, then
$$\lim_{n \rightarrow \infty}\int_{a}^{b}f_{n}(x)dx = \int_{a}^{b}\lim_{n \rightarrow \infty}f_{n}(x)dx$$
It would seem like the theorem applies with $M = 1$, however, we know the above result is not true because $f$ is not integrable. Why does Dominated Convergence Theorem not apply to $f_{n}$?
Dominated convergence theorem of Riemann-integral is only valid when the limit function $f$ is assumed Riemann-integrable (the limit function will automatically be Lebesgue-integrable under the DCT-conditions).