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

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    $\begingroup$ What makes you think $f$ will be Riemann-integrable? $\endgroup$ – Alex Becker May 25 '12 at 12:30
  • $\begingroup$ Since $f_{n}$ is bounded and converges to $f$ pointwise. $\endgroup$ – Hassan Muhammad May 25 '12 at 12:33
  • $\begingroup$ In general you need uniform convergence for the function to remain Riemann-integrable. $\endgroup$ – Alex Becker May 25 '12 at 12:40
  • $\begingroup$ This paper is relevant. $\endgroup$ – Antonio Vargas May 25 '12 at 14:57
  • $\begingroup$ 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. $\endgroup$ – Arturo Magidin May 26 '12 at 1:52
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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.

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  • $\begingroup$ Yes, courses my life becomes easy now. Thanks@ncmathsadist. $\endgroup$ – Hassan Muhammad May 25 '12 at 12:37
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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.

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

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    $\begingroup$ Where can I find a proof of this? $\endgroup$ – Twink Dec 13 '13 at 5:25
  • $\begingroup$ @Twink see the article mentioned in the answer by M. Mueger. $\endgroup$ – KCd Oct 24 '18 at 1:23
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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

$$\int_{a}^{b}f_{n}(x)dx=\int_{E}f_{n}(x)dx+\int_{F}f_{n}(x)dx\leq K m(E_{n})+\frac{\epsilon}{2(b-a)} (b-a)\leq \epsilon.$$

$\square$

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