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Jan 29, 2003, 10:55:28 PM1/29/03

to

Why is it true that f is Riemann integrable on [a,b] with f : [a,b] --> R if

and only if {x is in [a,b] s.t. f is discontinuous at x} has Lebesgue

measure zero?

and only if {x is in [a,b] s.t. f is discontinuous at x} has Lebesgue

measure zero?

-Steve

Jan 30, 2003, 11:24:10 AM1/30/03

to

This is best answered by reading a proof in a book, which is too messy to

post in ASCII. The proof of this fact is in probably any undergraduate

analysis book. It's theorem 10.33(b) in my copy of Rudin, _Principles of

Mathematical Analysis_, but my copy is an old Russian edition, so the

reference may vary. The proof is subtle enough that people have

(unsuccessfully) questioned its validity in times not so long ago.

>

> -Steve

>

>

>

======================================================================

Michael A. Van Opstall

Padelford C-113

ops...@math.washington.edu

http://www.math.washington.edu/~opstall/

Jan 30, 2003, 4:00:25 PM1/30/03

to

In article <b1bf70$ab00$1...@netnews.upenn.edu>,

f also must be bounded.

Basically the proof of the "if" goes like this.

Suppose |f| <= B on [a,b], and you want a partition of [a,b] on which

any two Riemann sums will differ by at most epsilon. For convenience

I'll take B = 1 and [a,b] = [0,1]. Let U be an open

set containing the set where f is discontinuous, such that

m(U) < epsilon/10. Let K = [0,1] \ U. f is continuous at each point p of

K, so there is an open interval around p on which

|f(x) - f(p)| < epsilon/10 (and therefore |f(x) - f(y)| < epsilon/5 for

any x,y in this interval). Since K is compact and covered by these

open intervals, there is a finite subcover: a finite collection F_1 of

open intervals whose union V contains K. Then [0,1] \ V is the union of

finitely many closed intervals (some of which may be single points), with

total length < epsilon/10. Replacing each of these by a slightly larger

open interval, we get a finite collection F_2 of open intervals of total

length < epsilon/5. Together, F_1 and F_2 cover [0,1]. Now if you take

a fine enough partition, every interval of the partition will be contained

in a member of F_1 or F_2. Estimate, for any two Riemann sums on such

a partition, the difference of the contributions from the intervals

contained in members of F_1 and similarly for F_2, and you should get

that the total difference is less than epsilon.

For the "only if", note that if you have a partition such that any two

Riemann sums differ by less than, say, epsilon, the total length of the

intervals of the partition on which f varies by more than delta must

be at most epsilon/delta. Take epsilon -> 0, and it implies that the

measure of the set of x such that

lim sup_{y -> x} f(y) - lim inf_{y -> x} f(y) > delta

is 0. But the set of points where f is not continuous is the union of

a countable family of these sets, so it has measure 0.

Robert Israel isr...@math.ubc.ca

Department of Mathematics http://www.math.ubc.ca/~israel

University of British Columbia

Vancouver, BC, Canada V6T 1Z2

Jan 30, 2003, 4:22:58 PM1/30/03

to

On Thu, 30 Jan 2003 08:24:10 -0800, Michael van Opstall

<ops...@math.washington.edu> wrote:

<ops...@math.washington.edu> wrote:

>On Wed, 29 Jan 2003, Steve wrote:

>

>> Why is it true that f is Riemann integrable on [a,b] with f : [a,b] --> R if

>> and only if {x is in [a,b] s.t. f is discontinuous at x} has Lebesgue

>> measure zero?

>>

>

>This is best answered by reading a proof in a book, which is too messy to

>post in ASCII. The proof of this fact is in probably any undergraduate

>analysis book. It's theorem 10.33(b) in my copy of Rudin, _Principles of

>Mathematical Analysis_, but my copy is an old Russian edition, so the

>reference may vary. The proof is subtle enough that people have

>(unsuccessfully) questioned its validity in times not so long ago.

Really? This strikes me as odd. The proof takes a few lines

to write down, but it's not that hard (Robert Israel managed to

give a pretty complete sketch in a usenet post just now)

and actually it seems like the result is almost obvious when

you look at it right.

Well that may be something of an exaggeration. But

who questioned the validity of this result?

>> -Steve

>>

>>

>>

>

>======================================================================

>Michael A. Van Opstall

>Padelford C-113

>ops...@math.washington.edu

>http://www.math.washington.edu/~opstall/

David C. Ullrich

Jan 30, 2003, 4:45:50 PM1/30/03

to

>

>Why is it true that f is Riemann integrable on [a,b] with f : [a,b] --> R if

>and only if {x is in [a,b] s.t. f is discontinuous at x} has Lebesgue

>measure zero?

>

Because if g is the lower envelope of f and h is the upper envelope of f then f>Why is it true that f is Riemann integrable on [a,b] with f : [a,b] --> R if

>and only if {x is in [a,b] s.t. f is discontinuous at x} has Lebesgue

>measure zero?

>

is continuous at x iff h(x)=g(x). But f is Riemann integrable iff h-g=0 a.e.

QED. (See Royden's Real Analysis Ch 4.)

Rich Burge

Jan 30, 2003, 4:48:58 PM1/30/03

to

On Thu, 30 Jan 2003, David C Ullrich wrote:

> On Thu, 30 Jan 2003 08:24:10 -0800, Michael van Opstall

> <ops...@math.washington.edu> wrote:

>

> >On Wed, 29 Jan 2003, Steve wrote:

> >

> >> Why is it true that f is Riemann integrable on [a,b] with f : [a,b] --> R if

> >> and only if {x is in [a,b] s.t. f is discontinuous at x} has Lebesgue

> >> measure zero?

> >>

> >

> [my reply snipped]

>

> Really? This strikes me as odd. The proof takes a few lines

> to write down, but it's not that hard (Robert Israel managed to

> give a pretty complete sketch in a usenet post just now)

> and actually it seems like the result is almost obvious when

> you look at it right.

>

> Well that may be something of an exaggeration. But

> who questioned the validity of this result?

>

Maybe that was an irresponsible comment, since I can't back it up. I was

told when I was an undergraduate in real analysis that someone claimed to

have a proof to the contrary, but that this was someone regarded as a

crackpot...

Jan 31, 2003, 7:10:00 AM1/31/03

to

On Thu, 30 Jan 2003 13:48:58 -0800, Michael van Opstall

<ops...@math.washington.edu> wrote:

<ops...@math.washington.edu> wrote:

>On Thu, 30 Jan 2003, David C Ullrich wrote:

>

>> On Thu, 30 Jan 2003 08:24:10 -0800, Michael van Opstall

>> <ops...@math.washington.edu> wrote:

>>

>> >On Wed, 29 Jan 2003, Steve wrote:

>> >

>> >> Why is it true that f is Riemann integrable on [a,b] with f : [a,b] --> R if

>> >> and only if {x is in [a,b] s.t. f is discontinuous at x} has Lebesgue

>> >> measure zero?

>> >>

>> >

>> [my reply snipped]

>>

>> Really? This strikes me as odd. The proof takes a few lines

>> to write down, but it's not that hard (Robert Israel managed to

>> give a pretty complete sketch in a usenet post just now)

>> and actually it seems like the result is almost obvious when

>> you look at it right.

>>

>> Well that may be something of an exaggeration. But

>> who questioned the validity of this result?

>>

>

>Maybe that was an irresponsible comment, since I can't back it up. I was

>told when I was an undergraduate in real analysis that someone claimed to

>have a proof to the contrary, but that this was someone regarded as a

>crackpot...

That's different. When you say "The proof is subtle enough that

people have (unsuccessfully) questioned its validity in times not

so long ago" it sounds like there's some subtlety that might lead

a rational mathematician to question the proof's validity. If

you're just talking about _someone_ questioning the validity

that says nothing. Read sci.math for a while - you'll find

crackpots questioning the validity of all sorts of things,

none of which are the least bit "subtle".

>> >> -Steve

>> >>

>> >>

>> >>

>> >

>> >======================================================================

>> >Michael A. Van Opstall

>> >Padelford C-113

>> >ops...@math.washington.edu

>> >http://www.math.washington.edu/~opstall/

>>

>>

>> David C. Ullrich

>>

>

>======================================================================

>Michael A. Van Opstall

>Padelford C-113

>ops...@math.washington.edu

>http://www.math.washington.edu/~opstall/

David C. Ullrich

Jan 31, 2003, 11:58:26 AM1/31/03

to

In article <Pine.LNX.4.44.03013...@zeno2.math.washington.edu>,

>On Wed, 29 Jan 2003, Steve wrote:

>> Why is it true that f is Riemann integrable on [a,b] with f : [a,b] --> R if

>> and only if {x is in [a,b] s.t. f is discontinuous at x} has Lebesgue

>> measure zero?

>This is best answered by reading a proof in a book, which is too messy to

>post in ASCII. The proof of this fact is in probably any undergraduate

>analysis book. It's theorem 10.33(b) in my copy of Rudin, _Principles of

>Mathematical Analysis_, but my copy is an old Russian edition, so the

>reference may vary. The proof is subtle enough that people have

>(unsuccessfully) questioned its validity in times not so long ago.

The proof is straightforward enough, provided that

boundedness is added. Let A_e be the set of all

discontinuities large than size e. Observe that the

closure of A_e is contained in A_{e/2}. A bounded function

is Riemann integrable if and only if for any positive c the

interval can be divided into finitely many subintervals in

each of which the function varies by less than c, with the

total length left over being less than c. Now use the fact

that a closed set covered by a countable union of intervals

is covered by finitely many.

--

This address is for information only. I do not claim that these views

are those of the Statistics Department or of Purdue University.

Herman Rubin, Deptartment of Statistics, Purdue University

hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558

Jan 31, 2003, 5:29:03 PM1/31/03

to

Michael van Opstall <ops...@math.washington.edu>

[sci.math Jan 30 2003 6:00:36:000PM]

http://mathforum.org/discuss/sci.math/m/478399/478526

[sci.math Jan 30 2003 6:00:36:000PM]

http://mathforum.org/discuss/sci.math/m/478399/478526

wrote regarding the Riemann integrability continuity condition:

> Maybe that was an irresponsible comment, since I can't back it

> up. I was told when I was an undergraduate in real analysis

> that someone claimed to have a proof to the contrary, but

> that this was someone regarded as a crackpot...

Let's go back in history, well before Lebesgue [3, p. 254],

Vitali [6] [7], and Young [8] independently proved that a

necessary and sufficient condition for the Riemann integrability

of a bounded function is that its discontinuities form a set

of measure zero.

Riemann [5, available on-line] introduced his integral in 1854.

In this same paper he gave an example, correctly verified, of a

Riemann integrable function whose discontinuities form a dense

set. Riemann's paper wasn't generally known until it was published

after his death (in 1866) by Dedekind in 1868. Two years later,

Hankel [2, available on-line] proved that every Riemann integrable

function has a dense set of continuity points. (For more about

Hankel's result, see Renfro [4].)

Thus, for historical purposes, within TWO years after Riemann

gave his definition of the integral, it was known that Riemann

integrable functions are very well behaved and it was known

that they can be very badly behaved. They are so close to

being continuous that you'd think they can't be any worse

behaved than functions such as piecewise continuous functions,

the characteristic function of {1/2, 1/3, 1/4, ...}, etc. On

the other hand, they can be so badly behaved that it isn't

intuitively evident to a novice that such a function can be

continuous anywhere, except maybe at isolated points or on

sets such as {1/2, 1/3, 1/4, ...}, etc. [Let g(x) be the

characteristic of the rationals and consider functions such

as x*g(x) + (x-3)*g(x).]

I'd bet that whatever it is about the integrability condition

that this person was bothered with, their difficulty lies within

these two extremes -- extremes that were essentially known within

two years after Riemann's definition of the integral was published.

Incidentally, the definition that Riemann gave for his integral

is a minor modification of Cauchy's definition from 1823. Cauchy

used partitions whose associated subintervals have varying lengths,

but Cauchy used only left (or right) endpoints of these subintervals

to evaluate the function at. [Cauchy's seemingly more restrictive

definition actually gives rise to the same collection of integrable

functions that Riemann's definition does -- see Gillespie [1].]

Riemann's nontrivial contributions on this topic were (a) giving a

necessary and sufficient condition for integrability based on the

behavior of a function, (b) using this condition to prove the

integrability of a function having a dense set of discontinuities,

and (c) putting the focus on the collection of functions that are

integrable according to some notion of integrability, rather than

defining a notion of integrability only in order to rigorously

prove certain desired integrability properties. [I believe this

was the first time a function both continuous and discontinuous

on a dense set was given by anyone. A more familiar example is

the "ruler function", equal to 0 at the irrationals and 1/q for

rationals p/q in reduced form. The ruler function first appeared

in a paper by Thomae in 1875.)]

Cauchy, on the other hand, only proved that functions with finitely

many discontinuities in every interval are integrable. Moreover,

these were the only functions that Cauchy was interested in,

and so Cauchy never sought to generalize his results to more

pathological functions nor did he consider his definition of the

integral as defining a collection of functions that might be worthy

of study.

The Lebesgue/Vitali/Young integrability condition is actually a

small step away from results already known by 1875. The integrability

condition that Riemann gave involved the oscillation of a function

in an interval. Hankel [2] reformulated Riemann's condition in terms

of the oscillation of a function at a point, a notion Hankel also

came up with in this paper. Hankel's reformulated condition is a

bit closer to the notion of continuity at a point, since (as Hankel

was well aware) it is straightforward to show that f is continuous

at x=a if and only if the oscillation of f at x=a is zero.

Specifically, Hankel said that a function f is Riemann integrable

if and only if for each epsilon > 0 the set of points at which f

has an oscillation greater than epsilon can be covered by a finite

collection of intervals whose lengths have an arbitrarily small

sum. [This differs from the way Lebesgue measure zero sets are

defined in that only finitely many intervals are allowed for the

coverings.] Hankel's claim is true. However, Hankel's proof that

a Riemann integrable function has this property was incorrect.

[The other half of Hankle's "iff" claim, that this property implies

Riemann integrability, was correctly proved by Hankel.] Hankel's

error was that he tried to prove this by proving a stronger result,

namely that every function whose points of continuity form a dense

set will be Riemann integrable. Unfortunately, this last result is

too strong -- it's not even true! If it had been true, then Hankel's

proof would at least be correct at the top level, since every

Riemann integrable function has a dense set of continuity points.

However, the characteristic function of a Cantor set with positive

measure is a counterexample. [It was precisely to show this error

in Hankel's paper that H.J.S. Smith used a Cantor set of positive

measure in his 1875 paper that contained the first ever construction

of Cantor sets (whether of measure zero or of positive measure, and

Smith gave both kinds).]

Hankel's proof was repaired by Ascoli in 1875. An even more precise

version was given by Volterra in 1881. We say that a set E is

"Jordan null" (or has Jordan content zero) if, given any delta > 0,

there exists a finite collection of intervals covering E whose

lengths have a sum less than delta. Then, using this notion, which

I might add is equivalent (if E is bounded) to the closure of E

having Lebesgue measure zero), Hankel's integrability condition

says that f is Riemann integrable if and only if for each epsilon > 0

the points at which f has an oscillation greater than epsilon form

a Jordan null set. Volterra showed that the result continues to hold

if we use the "left oscillation of f" rather than the oscillation

of f, or equivalently (consider -f), the right oscillation of f.

[Left and right oscillation at a point are defined in the obvious

way.]

The jump from these results to Lebesgue's result is small once

the notion of a set of measure zero is available, since it's

essentially a matter of a countable union of measure zero sets

having measure zero. To be explicit, we can obtain a covering

of the discontinuities of f with intervals whose lengths have

a sum less than delta by taking the union over n = 1, 2, 3, ...

of the following finite collections C_n of intervals that we

know exist by Hankel's condition -- Given n, let C_n be a

finite collection of intervals covering the points where f

has an oscillation greater than 1/n and such that the sum of

the lengths of the intervals in C_n is less than delta / 2^n.

[1] D. C. Gillespie, "The Cauchy definition of a definite integral",

Annals of Mathematics (2) 17 (1915-16), 61-63. [JFM 45.0441.02]

http://www.emis.de/cgi-bin/JFM-item?45.0441.02

[2] Hermann Hankel, "Untersuchungen über die unendlich oft

oscillirenden und unstetigen functionen", Math. Ann. 20 (1882),

63-112. [JFM 14.0320.02] [Originally presented at the Univ. of

Tübingen on March 6, 1870 (JFM 02.0190.01). A French translation

appears in "Cahiers du séminaire d'histoire des mathématiques",

edited by P. Dugac and R. Taton, 9, Univ. Paris VI, 139-209

(MR 89f:01090).]

http://www.emis.de/cgi-bin/JFM-item?02.0190.01

http://www.emis.de/cgi-bin/JFM-item?14.0320.02

http://134.76.163.65:80/agora_docs/35867TABLE_OF_CONTENTS.html

[3] Henri Lebesgue, "Intégrale, longueur, aire", Annali di

Mathematica pura ed applicata (3) 7 (1902), 231-359.

[JFM 33.0307.02]

http://www.emis.de/cgi-bin/JFM-item?33.0307.02

[4] Dave L. Renfro, Historia-Matematica post, January 28, 2003.

http://mathforum.org/epigone/historia_matematica/bolkyrland/k95z674emvvz@legacy

[5] Bernhard Riemann, "Ueber die Darstellbarkeit einer Function

durch eine trigonometrische Reihe", Habilitationsschrift thesis,

1854. [Published posthumously in Abhandlungen der Königlichen

Gesellschaft der Wissenschaften zu Göttingen 13 (1868), 87-131

(JFM 01.0131.03).]

http://www.emis.de/cgi-bin/JFM-item?01.0131.03

http://www.maths.tcd.ie/pub/HistMath/People/Riemann/Trig/

[6] Giuseppe Vitali, "Sulla condizione di integrabilità delle

funzioni", Bullettino Mensile della Accademia Gioenia di

Scienze Naturali in Catania 79 (1903), 27-30. [JFM 34.0417.03]

http://www.emis.de/cgi-bin/JFM-item?34.0417.03

[7] Giuseppe Vitali, "Sulla integrabilità delle funzioni", Rendiconti

Reale Instituto Lombardo di Scienze e Lettere (2) 37 (1904),

69-73. [JFM 35.0393.02]

http://www.emis.de/cgi-bin/JFM-item?35.0393.02

[8] William H. Young, "A note on the condition of integrability

of a function of one real variable", The Quarterly Journal of

Pure and Applied Mathematics 35 (1904), 189-192. [JFM 34.0417.01]

http://www.emis.de/cgi-bin/JFM-item?34.0417.01

Dave L. Renfro

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