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In other words, how to prove

A continuous function over a closed interval is Riemann-integrable. That is, if a function $f$ is continuous on an interval $[a, b]$, then its definite integral over $[a, b]$ exists.

Edit:

The Definite Integral as a Limit of Riemann Sums :

Let $f(x)$ be a function defined on a closed interval $[a, b]$. We say that a number $I$ is the definite integral of $f$ over $[a, b]$ and that $I$ is the limit of the Riemann sums $\sum \limits_{k=1}^n f(c_k)\Delta x_k$ if the following condition is satisfied:

Given any number $\epsilon \gt 0$, there is a corresponding number $\delta \gt 0$ such that for every partition $P = \{x_0, x_1, ... , x_n\}$ of $[a, b]$ with $\|P \| < \delta$ and any choice of $c_k$ in $[x_{k-1}, x_k]$, we have $$ \left| \sum_{k=1}^n f(c_k) \Delta x_k - I \ \right| \lt \epsilon .$$

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In the Riemann sense? Which definition of the Riemann integral are you using? –  Qiaochu Yuan Aug 8 '11 at 22:23
    
Read wikipedia article. When writing out Riemann sum, approximate your function by a constant on that interval, convince yourself of convergence. –  Sasha Aug 8 '11 at 22:25
    
You're not working with a definition it's easy to prove things about (since you need to exhibit $I$ to verify it). It's much easier to use the Darboux integral, and there the proof is fairly straightforward and a nice exercise: en.wikipedia.org/wiki/Darboux_integral –  Qiaochu Yuan Aug 8 '11 at 22:54
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Yeah, I know the definition made the proof difficult. That's why I asked the question. –  ablmf Aug 8 '11 at 23:03
    
@Qiaochu: I can be characterized as $g(b)-g(a)$ for an an antiderivative $g$ of $f$. I believe that this would lead quite quickly to a proof using the above definition of the integral. However, one would then need to establish the fact that continuous functions are indefinitely integrable without appealing to the Riemann integral (and the FTC), and I don't see at the moment how to do this. –  Mark Aug 8 '11 at 23:31
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3 Answers 3

up vote 8 down vote accepted

First note that a precise formulation of your question is:

How do you prove that every continuous function on a closed bounded interval is Riemann (not Darboux) integrable?

You can find a proof in these notes, from an undergraduate analysis course I taught some years ago at McGill University.

Here is a rough outline of this handout:

I. I introduce the ("definite") integral axiomatically. One of the axioms is that the set of integrable functions on $[a,b]$ should contain all the continuous functions.
II. I prove that the Fundamental Theorem of Calculus follows (easily) from the axioms.
III. I introduce the notion of a Riemann integrable function (which is exactly what you wrote above) and verify that the class of Riemann integrable functions on $[a,b]$ satisfy the axioms of Part I. In particular:
IV: I prove that every continuous function is Riemann integral (Theorem 10, starting on page 7).

In the next handout I start talking about the Darboux integral and how it compares to the Riemann integral. But it was an intentional decision to present the Riemann integral first. This is what students are expecting from their previous courses, and it is not so bad to work with, at least for a while.

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Ha, I will an graduate student in McGill this September and I'm reviewing some math knowledge. I guess understanding the proof of this theorem is a bit out of my ability. Thanks anyway! –  ablmf Aug 10 '11 at 0:31
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Riemann-Darboux integral:

For the Riemann-Darboux integral it is easier than with your definition (which is equivalent in $\mathbf R$).

The function is uniformly continuous on $[a,b]$ (why?). This means that if $\epsilon > 0$ is given we can find $\delta = \delta(\epsilon) > 0$ such that $|x - y| < \delta$ implies $|f(x) - f(y)| < \frac{\epsilon}{2(b - a)}$. Now let $P_\epsilon$ be a partition with norm $\|P_\epsilon\| < \delta$. Now for $P$ finer than $P_\epsilon$ we have

$$M_k(f) - m_k(f) \leq \frac{\epsilon}{2(b - a)}.$$

Where $M_k$ is the supremum in $[x_{k - 1}, x_k]$ and $m_k$ is the infimum. Multiply this inequality with $\Delta x_k$ and sum to get

$$U(P, f) - L(P, f) \leq \frac{\epsilon}{2(b - a)} \sum_{k = 1}^n \Delta x_k = \frac\epsilon2 < \epsilon.$$

Fine, so what is left is to prove the same thing for your definition of the integral (not so easy) or proving the equivalence between the two (not so hard).

Riemann integral:

There is a problem with the above approach if we are in a general Banach space (why?), so we must resort to the normal Riemann integral.

Let $f:[a,b] \to E$ be continuous where $E$ is a Banach space. Given $\epsilon > 0$ let $\delta$ be such that $$\text{if } |x - y| < \delta \text{ then } |f(x) - f(y)| < \frac{\epsilon}{b - a}.$$

Let $P$ and $P'$ be partitions of $[a,b]$ with norm smaller than $\delta$. Let $c$ and $c'$ be the choices of points in each interval of $P$ and $P'$ respectively. We want to estimate $|S(P, c) - S(P', c')|$ where $S(P, c)$ is the Riemann sum associated with $P$ and $c$. WLOG let $P \subset P'$. (If $P = P'$ then $$|S(P, c) - S(P, c')| \leq \sum |f(c_i) - f(c_i')| \Delta x_i \leq \epsilon.$$

Now suppose that $P'$ is obtained from $P$ by inserting one point (split one interval) for example say we insert $x_j'$ with $x_j \leq x_j' \leq x_{j + 1}$. In this case the partition size will not increase. WLOG assume that for $i \neq j$ we have $x_i' = x_i$ and that $c_j = x_j'$ and that $x_j'$ is also selected as the points in the intervals $[x_j, x_j']$ and $[x_j', x_{j + 1}]$. Now $S(P, c) - S(P', c') = 0$. We can repeat this process for another refinement.

This will give us the result (why?).

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Don't you have to prove that $|S(P,c)-S(P',c')|<\epsilon$ for all $(P,c)$ and $(P',c')$ with $\|P\|<\delta$ and $\|P'\|<\delta$. Restricting to those for which either $P\subseteq P'$ or $P\subseteq P'$ might miss some partitions... –  Freeze_S Jun 9 at 21:12
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Appendix to Chapter 13 in: CALCULUS, by M. Spivak.
Yes, uniform continuity is the key.

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In the same book, M. Spivak a very amazing and marvelous proof without the use of uniform continuity. In case you don't have access to this book, you can read the same proof online at paramanands.blogspot.com/2012/07/… –  Paramanand Singh Jul 15 '13 at 3:29
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