Does dominated Covergence Theorem hold for Riemann integral?

Question

Statement of the DCT:

Theorem - Let $\{f_n\}$ be a sequence in $L^1$ such that (a) $f_n\rightarrow f$, and (b) there exists a nonnegative $g\in L^1$ such that $|f_n|\leq g$ a.e. for all $n$. Then $f\in L^1$ and $\int f = \lim_{n\rightarrow \infty}\int f_n$.

Consider $(a_n)$ an enumeration of the rationals over $[0,1]$ and $f_n(x) = \sum \mathbf 1_{{x = a_n}}$. — nicomezi, Jun 01 '20 at 19:32

score 0 · Answer 1 · answered Jun 01 '20 at 19:31

0

It depends. It is possible that $f$ is not Riemann-integrable but is Lebesgue integrable, so you should interpret $\int f$ as a Lebesgue-integral for this to hold.

answered Jun 01 '20 at 19:31

J. De Ro

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Does dominated Covergence Theorem hold for Riemann integral?

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