Lebesgue Measure zero
Steve
and only if {x is in [a,b] s.t. f is discontinuous at x} has Lebesgue
measure zero?
-Steve
Michael van Opstall
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/
Robert Israel
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
David C Ullrich
<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
R3769
>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
Michael van Opstall
> 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...
David C Ullrich
<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
Herman Rubin
>> 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
Dave L. Renfro
[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