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Definitions

Suppose $P\colon a=x_0<x_1<\dotsb<x_n=b$ is a partition of $[a,b]$. Let $\Delta x_k=x_k-x_{k-1}$ and $\lVert P\rVert$ denotes $\max_{0<k\le n}\Delta x_k$.

The Cauchy integral of a function $f$ on closed interval $[a,b]$ equals to $I$ if and only if for each $\epsilon>0$, there's some $\delta>0$, for each partition $P$ of $[a,b]$ such that $\lVert P\rVert<\delta$, we have $\left\lvert\sum_{k=1}^nf(x_k)\Delta x_k-I\right\rvert<\epsilon$.

Problem

If $f$ is bounded on $[a,b]$ whose Cauchy integral equals to $I$, then $f$ is Riemann-integrable and $\int_a^bf=I$.

Background

It's an exercise from our calculus(analysis) problemset book, and there's a hint: consider the partitions whose $x_k-x_{k-1}$ is a constant for different $k$'s, and try to estimate the Riemann sum for each of these partitions through the Cauchy integral.

I have no idea about such estimation. After drawing some pictures, I discouraged. I googled on the Internet and found an article. I realized that it's a quite different approach and with some advanced techniques (such as the analysis of a positive measure set -- discontinuities). I hope there will be some simpler approachers, just as the hint says. I need a more detailed hint, or a solution. Can anybody help me? Thanks!

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What book are you using to study this? –  Jayesh Badwaik Mar 14 '13 at 22:51
    
@JayeshBadwaik A reference book that hasn't been translated into English, but I think the problems are assembled from some other books. –  Frank Science Mar 16 '13 at 13:07
    
It will be better to add the definition of Riemann integral you using. –  leo Apr 7 '13 at 6:21
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$@$Frank Science: books which have not been translated into English still have titles and bibliographic information. In my opinion if you are asking for help with something from a book you're reading and someone asks you what the book is, the courteous thing to do would be to tell them. –  Pete L. Clark Apr 7 '13 at 8:47

3 Answers 3

up vote 2 down vote accepted
+25

D.C.Gillespie proved the theorem in 1915 (Annals of Mathematics, Vol.17) and what a proof !
To propose the proof as an exercise in a calculus book seems rather strange ...

However see exercise 2.1.19 in Bressoud's A Radical Approach to Lebesgue's Theory of Integration. There is a hint on page 300. Can it help ?

See also theorem 1 in Kristensen, Poulsen, Reich A characterization of Riemann-Integrability, The American Mathematical Monthly, vol.69, No.6, pp. 498-505.

But the story is the same !

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Just a historical comment: in 1915 (and before) the Annals was not the tip-top mathematical journal that it is today. I presume it was the top (and one of relatively few) American journals, but American mathematics back then lagged far behind European mathematics. See e.g. math.uga.edu/~pete/Miller03.pdf for an Annals paper that looks undergraduate level by present standards. (Which is not to say that Gillespie's paper is easy: I glanced at it, and I don't find it to be!) –  Pete L. Clark Apr 7 '13 at 8:53
    
Gillespie's is just the one I've mentioned in my text by hyperlink article. After some efforts, now I believe that the elementary proof of squeezing Riemann sums through Cauchy sums would be meaningless, even though there were. –  Frank Science Apr 12 '13 at 16:00

I think you should try to use the definition of the Riemann integral and work with the Darboux sums: here.

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The Darboux integral is not the same as the Cauchy integral. You still have to show that those two coincide –  Ewan Delanoy Apr 1 '13 at 7:57
    
I agree, but by using the characterization of the Cauchy integral in the definition of the problem we almost already have the equivalence between both of them. –  Nre Apr 1 '13 at 10:59
    
There are two sides to the equivalence, one very easy as you say and the other not obvious at all. It is very easy to control a Cauchy sum between two Riemann/Darboux sums (in fact, a Cauchy sum nothing but a special Riemann sum). What is not clear is whether we can control a Riemann/Darboux sum between two Cauchy sums, and this is precisely what the original question is about. –  Ewan Delanoy Apr 1 '13 at 11:32
    
@EwanDelanoy Haha, yeah. I found it somewhat easier to follow the way of that article. Suppose that $f$ is bounded. If $f$ is not Riemann-integrable, we can show that there's some $\epsilon_0,\eta_0>0$ such that the set of points whose oscillations $\ge\epsilon_0$, say $D$, is compact and total length of any open covering $>\eta_0$, then we can construct two Cauchy-sums differs a lot with arbitrary small $\|P\|$. Unfortunately, I haven't found a direct way to estimate Darboux sums through Cauchy sums. –  Frank Science Apr 3 '13 at 15:03
    
@FrankScience : indeed, perhaps there isn’t. In that case, it might be very hard to show there isn’t. –  Ewan Delanoy Apr 3 '13 at 15:05

I think for a calculus class/book, this might be overcomplicated. Though, before having a proper measure theory course, I was taught that a function is integrable if it has a limited number of discontinuities (or class I and II discontinuities), apart from being bounded.

$$\sum_{k=1}^{n}\left | f(t_{k})\cdot \bigtriangleup x_{k} - I \right | = \sum_{k=1}^{n}\left | f(t_{k})\cdot \bigtriangleup x_{k} - f(x_{k})\cdot \bigtriangleup x_{k} + f(x_{k})\cdot \bigtriangleup x_{k} - I \right | \leq \sum_{k=1}^{n} \left \{ \left | f(t_{k}) - f(x_{k}) \right |\cdot \bigtriangleup x_{k} + \left | f(x_{k})\cdot \bigtriangleup x_{k} - I \right | \right \} \leq \varepsilon + \sum_{k=1}^{n} \left | f(t_{k}) - f(x_{k}) \right |\cdot \bigtriangleup x_{k} $$

Now, if $f(x)$ is continuous, then it is uniform continuous on $[a,b]$ so that we can pick up $P$ such that $max \left | f(t_{k}) - f(x_{k}) \right | \leq \frac{\varepsilon }{b-a}$ and then: $$\sum_{k=1}^{n} \left | f(t_{k}) - f(x_{k}) \right |\cdot \bigtriangleup x_{k} \leq \frac{\varepsilon }{b-a} \sum_{k=1}^{n} \bigtriangleup x_{k} = \varepsilon$$

If $f(x)$ is bounded and has a limited number of discontinuities, then we can split $[a,b]$ into sub-segments and "make" $f(x)$ continuous on those sub-segments. It should be enough to prove this for the case with one discontinuity, and use induction after that.

However, the necessary and sufficient condition states that the set of points of discontinuity should have measure of zero. If it is so, then: $$\sum_{k=1}^{n} \left | f(t_{k}) - f(x_{k}) \right |\cdot \bigtriangleup x_{k}$$ can be split into: $$\sum_{k=1}^{n} \left | f(t_{k}) - f(x_{k}) \right |\cdot \bigtriangleup x_{k} \leq M \cdot \mu(D_{n-m}) + \sum_{i=1}^{m} \left | f(t_{k_{i}}) - f(x_{k_{i}}) \right |\cdot \bigtriangleup x_{k_{i}} $$ where summation is on the continuous part of the $[a,b]$, $M=max|f(x)|$, $D_{n-m}$ is the remaining part of $[a,b]$ containing the set of points of discontinuity and $\mu(D_{n-m})$ is its measure which goes to zero. So, this is the biggest question; how, using the fact that Cauchy integral is equal to $I$, to prove that the set of points of discontinuity has the measure of zero? I tend to agree with @Tony Piccolo answer ...

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Let $\omega(x_0)=\lim_{u\to0}(\sup f(x)-\inf f(x))$ be the oscillation around $x_0$, we have the discontinuities $D=\cup_n D_n$ where $D_n=\{x\colon\omega(x)\ge 1/n\}$ are compact subsets, therefore $m(D)=0$ if and only if $m(D_n)=0$ for all $n$, and note that $m(D_n)$ is just the content, not only the measure. The characterization about $D_n$, of Riemann-integrability for bounded functions, which could be proved elementarily, is more and more prevalent on the calculus books. –  Frank Science Apr 12 '13 at 16:11

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