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 $$\int_{E}f=\lim_{n \to \infty}\int_{E}f_{n}.$$

 A: 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.
A: 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 $$\lim\limits_{n\to\infty}\int_a^bf_n(x)\,dx = \int_a^bf(x)\,dx$$

A: 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.
A: I typed the proof in Proofwiki following: Lewin, Jonathan W. "A truly elementary approach to the bounded convergence theorem." The American Mathematical Monthly 93.5 (1986): 395-397
== Lemma==
We call $E\subset \mathbb{R}$ an elementary subset if $E=\bigcup_{k=1}^{M} [a_{k},b_{k}]$ and we define $m(E)$ as the total length of these intervals minus their overlaps.
Lemma: Suppose $(A_{n})$ is a contracting sequence of bounded sets in R with an empty intersection. Let $$a_{n}:=\sup\{m(E): E\subset A_{n}\text{ is an elementary subset}  \}.$$
Then $a_{n}\to 0$.
== Proof of lemma==
The sequence $a_{n}$ is decreasing and assume that $a_{n}\geq \delta>0$ to obtain a contradiction. 
By the epsilon definition of supremm, for $\epsilon:=\frac{\delta}{2^{n}}$ there exists elementary subset $E_{n}$ such that
$$ m(E_{n})\geq a_{n}-\frac{\delta}{2^{n}}.$$
For $H_{n}=\bigcap_{k=1}^{n}E_{k}\subset \bigcap_{k=1}^{n}A_{k}$, we will show that $H_{n}\neq \varnothing$ and thus contradict that $A_{n}$ have an empty intersection.
For each n, take any elementary subset $E\subset A_{k}\setminus E_{k}$, then we find 
$$m(E)+m(E_{k})=m(E\cup E_{k})\leq a_{k}\Rightarrow m(E)\leq \frac{\delta}{2^{k}}.$$
Now take an elementary subset $S\subset A_{n}\setminus H_{n}=\bigcap_{k=1}^{n}(A_{n}\setminus E_{k})$, then we find
$$E=(E\setminus E_{1})\cup … \cup (E\setminus E_{n}).$$
Therefore, we get the bound
$$m(E)\leq \sum_{k=1}^{n}m(E\setminus E_{k})\leq  \sum_{k=1}^{n}\frac{\delta}{2^{n}}=\delta.$$
In words, any elementary subset $E\subset A_{n}\setminus H_{n}$ was shown to have measure $m(E)\leq \delta$. 
However, the inequality $a_{n}>\delta$ requires the existence of at least one elementary subset $U_{n}\subset A_{n}$ s.t. $m(U_{n})>\delta$. 
Since all the elementary subset $E\subset A_{n}\setminus H_{n}$ satisfy $m(E)\leq \delta$, we must have that $U_{n}\subset H_{n}$ for $n\geq 1$. 
This contradicts the non-emptiness because $\lim_{n\to \infty} m(U_{n})>\delta$.
$\square$
== Proof of main result==
WLOG assume that $f_{n}\geq 0$ and $f_{n}\to 0$, so we will show that given $\epsilon>0$ there exists N s.t.  forall $n\geq N$ we have . 
$$\int_{a}^{b}f_{n}(x)dx\leq \epsilon.$$
Let $A_{n}:=\{x\in [a,b]:\text{ there exists }k\geq n \text{ such that} f_{k}(x)\geq \frac{\epsilon}{2(b-a)} \}$.
These sets are decreasing as $n\to +\infty$ and have empty intersection and so the sup $a_{n}$ from above goes to zero $a_{n}\to 0$. 
So let $E_{n}\subset A_{n}$ be an elementary subset with $m(E_{n})\leq \frac{\epsilon}{2K}$ for all $n\geq N$, and consider the following subsets
$$E:=\{x\in E_{n}:\text{ there exists }k\geq n \text{ such that} f_{k}(x)\geq \frac{\epsilon}{2(b-a)} \}\text{ and } F:=[a,b]\setminus E.$$
Therefore, we find
$$\int_{a}^{b}f_{n}(x)dx=\int_{E}f_{n}(x)dx+\int_{F}f_{n}(x)dx\leq K m(E_{n})+\frac{\epsilon}{2(b-a)} (b-a)\leq \epsilon.$$
$\square$
