# 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? May 25, 2012 at 12:30
• Since $f_{n}$ is bounded and converges to $f$ pointwise. May 25, 2012 at 12:33
• In general you need uniform convergence for the function to remain Riemann-integrable. May 25, 2012 at 12:40
• May 25, 2012 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. May 26, 2012 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. May 25, 2012 at 12:37

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.

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?
– Twnk
Dec 13, 2013 at 5:25
• @Twink see the article mentioned in the answer by M. Mueger.
– KCd
Oct 24, 2018 at 1:23

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

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

$$\square$$