# Does Bounded Covergence Theorem hold for Riemann integral?

Just after studying the Bounded Convergence Theorem BCT for Lebesgue integral, I asked myself a question. Does the BCT hold for Riemann? I answered YES since the function is bounded according to the hypothesis of the BCT. But some Lebesgue integral are not Riemann, this is where I got confused, please I need a guide from experts in the field.

Thanks.

Statement of the BCT:

Let $\{f_{n}\}$ be a sequence of measurable functions defined on a set $E$ of finite measure. Assume $\{f_{n}\}$ converges to $f$ pointwise and also $\{f_{n}\}$ is bounded for all $n$. Then $$\int_{E}f=\lim_{n \to \infty}\int_{E}f_{n}.$$

-
What makes you think $f$ will be Riemann-integrable? – Alex Becker May 25 '12 at 12:30
Since $f_{n}$ is bounded and converges to $f$ pointwise. – Hassan Muhammad May 25 '12 at 12:33
In general you need uniform convergence for the function to remain Riemann-integrable. – Alex Becker May 25 '12 at 12:40
– Antonio Vargas May 25 '12 at 14:57
I believe the corresponding convergence theorem requires uniform convergence. See this post for a summary of "FTC" and "convergence theorems" for several different types of integrals. The role of dominated convergence for Lebesgue integrals is played by uniform convergence for Riemann integrals. – Arturo Magidin May 26 '12 at 1:52

No. Enumerate the rationals in [0,1] with the sequence $\{r_n\}_{n=1}^\infty$. Now define $f_n(x)$ by $f_n(x) = 1$ if $x = r_k$ for some $1\le k \le n$ and 0 otherwise. For all $n$, we have $$\int_0^1 f_n(x)\,dx = 0.$$ However, the limit function, the indicator of the rationals in $[0,1]$ is not Riemann integrable. The bounded convergence theorem fails for the Riemann Integral.

-
Yes, courses my life becomes easy now. Thanks@ncmathsadist. – Hassan Muhammad May 25 '12 at 12:37

But this statement is true:

Let $\{f_n\}$ be a sequence of Riemann Integrable functions such that $f_n:[a,b]\rightarrow\mathbb{R}$ and $|f_n(x)|<M$ for all $n\geq1$ with $M>0$. Suppose that $f_n\to f$ pointwise where $f:[a,b]\rightarrow\mathbb{R}$ is Riemann Integrable. Then $$\lim\limits_{n\to\infty}\int_a^bf_n(x)\,dx = \int_a^bf(x)\,dx$$

-
Where can I find a proof of this? – Twink Dec 13 '13 at 5:25

A dominated convergence theorem for the Riemann integral exists, due to Arzel`a. But one needs the addtional assumption that the limit function is Riemann integrable, since this does not follow from pointwise bounded convergence. For a proof see either W. A. J. Luxemburg: Arzela's Dominated Convergence Theorem for the Riemann Integral. The American Mathematical Monthly, Vol. 78, No. 9 (Nov., 1971), 970-979 or the book "An interactive introduction to mathematical analysis" by J. Lewin, Cambridge Univ. Press, 2003, 2014.

-