lusin's theorem
jennifer
which one would be ideal and how can i give an example?
The World Wide Wade
<1163447570.8...@h54g2000cwb.googlegroups.com>,
"jennifer" <scrilla...@yahoo.com> wrote:
> What would happen if i removed an assumption from lusins theorem and
> which one would be ideal and how can i give an example?
Please state Lusin's Theorem precisely.
Dave L. Renfro
Your question is a bit vague, but perhaps something
in the following post I made on July 16, 2005 will help:
http://groups.google.com/group/sci.math/msg/680691c6eeb50b91
Dave L. Renfro
jennifer
Lusin's Theorem:
Let f be a measurable real valued function on an interval [a,b].Then
given e>0, there is a continuous function t on [a,b] s.t. m{x:f(x) does
not equal t(x)}< e.
what if i let m{x:f(x) = t(x)}<e ?
I am trying to delete one of the assumptions in this theorem to show
where, why and how the conclusion of the theorem is false.
The World Wide Wade
<1163455562.4...@m7g2000cwm.googlegroups.com>,
"jennifer" <scrilla...@yahoo.com> wrote:
> Dave L. Renfro wrote:
> > jennifer wrote:
> >
> > > What would happen if i removed an assumption from
> > > lusins theorem and which one would be ideal and how
> > > can i give an example?
> >
> > Your question is a bit vague, but perhaps something
> > in the following post I made on July 16, 2005 will help:
> >
> > http://groups.google.com/group/sci.math/msg/680691c6eeb50b91
> >
> > Dave L. Renfro
>
> Lusin's Theorem:
>
> Let f be a measurable real valued function on an interval [a,b].Then
> given e>0, there is a continuous function t on [a,b] s.t. m{x:f(x) does
> not equal t(x)}< e.
>
> what if i let m{x:f(x) = t(x)}<e ?
That makes zero sense.
> I am trying to delete one of the assumptions in this theorem to show
> where, why and how the conclusion of the theorem is false.
Suppose you replace [a,b] by R. Is it still true?
Suppose f is not measurable. Is it still true?
Hint: One of the answers is yes, the other one is no.
W. Dale Hall
jennifer wrote:
... stuff deleted ...
>>jennifer wrote:
>>
>>
>>>What would happen if i removed an assumption from
>>>lusins theorem and which one would be ideal and how
>>>can i give an example?
>>
... stuff deleted ...
>
>
> Lusin's Theorem:
>
> Let f be a measurable real valued function on an interval [a,b].Then
> given e>0, there is a continuous function t on [a,b] s.t. m{x:f(x) does
> not equal t(x)}< e.
>
> what if i let m{x:f(x) = t(x)}<e ?
>
> I am trying to delete one of the assumptions in this theorem to show
> where, why and how the conclusion of the theorem is false.
>
Note that your suggestion here takes the *conclusion* of the theorem
and stands it on its head. This is a far cry from removing one of the
assumptions. WWW's suggestion is well worth following.
Dale.
jennifer
on R it would still work but with f not measurable it fails.
Is there an example that i can give of a function not measurable that
fails lusins criteria?
David C. Ullrich
wrote:
??? If you don't know an example then how do you know that it
fails for non-measurable f?
Anyway, a hint: Maybe you could even show that if
f satisfies the conclusion of Lusin's theorem then
f _must_ be measurable...
David C. Ullrich
Robert Israel
Of course she still won't know an example of a non-measurable
function. Nor will she ever, for some interpretations of "know
an example".
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
jennifer
sexist pig
cbr...@cbrownsystems.com
> Robert Israel wrote:
> > David C. Ullrich wrote:
<snip>
> > > Anyway, a hint: Maybe you could even show that if
> > > f satisfies the conclusion of Lusin's theorem then
> > > f _must_ be measurable...
> >
> > Of course she still won't know an example of a non-measurable
> > function. Nor will she ever, for some interpretations of "know
> > an example".
> >
> > Robert Israel isr...@math.ubc.ca
> > Department of Mathematics http://www.math.ubc.ca/~israel
> > University of British Columbia Vancouver, BC, Canada
>
> sexist pig
I think you misunderstand Robert's comment; he means that /no one/ can
"know an example" of a non-measurable function, just like no one can
"know an example" of a well-ordering of R.
Cheers - Chas
Dave L. Renfro
>> Of course she still won't know an example of a
>> non-measurable function. Nor will she ever, for
>> some interpretations of "know an example".
jennifer wrote:
> sexist pig
He's talking about set-theoretic definability issues relating
to things like the axiom of choice. Indeed, each nonmeasurable
function fails to belong to any of the Baire classes of
functions, meaning it's pretty much impossible to reach by
any sort of countably iterated limiting process.
http://mathworld.wolfram.com/BaireFunction.html
http://groups.google.com/groups?q=nonmeasurable+axiom-of-choice
Mirko Sardella, "What's the price of a nonmeasurable set"
Mathematica Bohemica 127 (2002), 41-48.
http://www.emis.de/journals/MB/127.1/mb127_1_5.pdf
Dave L. Renfro
jennifer
what if e= 0, as the assumption that i am deleting?
Dave L. Renfro
> what if e= 0, as the assumption that i am deleting?
Did you look at my sci.math essay on Lusin's theorem
that I cited in my first post in the present thread?
It discusses e = 0 and many related issues.
Dave L. Renfro
jennifer
But it does not tell where, how and why the conclusion of lusin theorem
fails.
In particular, when f is measurable on [a,b] is deleted.
jennifer
But it does not tell where, how and why the conclusion of lusin theorem
The World Wide Wade
<1163553529.1...@m73g2000cwd.googlegroups.com>,
"jennifer" <scrilla...@yahoo.com> wrote:
Clueless twit.
David C. Ullrich
wrote:
Of course - that's why I included the fact that she might
nonetheless be able to show that the conclusion of Lusin's
theorem actually implies that f is measurable.
>Robert Israel isr...@math.ubc.ca
>Department of Mathematics http://www.math.ubc.ca/~israel
>University of British Columbia Vancouver, BC, Canada
************************
David C. Ullrich
David C. Ullrich
wrote:
Idiot.
************************
David C. Ullrich
Dave L. Renfro
>>> what if e= 0, as the assumption that i am deleting?
Dave L. Renfro wrote:
>> Did you look at my sci.math essay on Lusin's theorem
>> that I cited in my first post in the present thread?
>> It discusses e = 0 and many related issues.
jennifer wrote:
> But it does not tell where, how and why the conclusion of
> lusin theorem fails.
>
> In particular, when f is measurable on [a,b] is deleted.
sci.math essay on Lusin's theorem
http://groups.google.com/group/sci.math/msg/680691c6eeb50b91
You asked me about e = 0. That's in my essay under REMARK 1.
Now you're asking about measurability of f. That's addressed
in my essay under REMARK 2. I didn't give a proof there, but
now that you know the result, you can look in real analysis
texts for it -- Bruckner/Brucker/Thomson (1997), Cohn (1980),
Foran (1991), Goffman (1953), Gordon (1994), Hann/Rosenthal (1948),
Hewitt/Stromberg (1965/1975), Natanson (1961), Oxtoby (1980),
Stromberg (1981), for example. [I got these from a LaTex
file of the paper I mentioned in my Lusin's theorem post.
I don't have any of them with me now, so I couldn't tell
you which ones have a proof of what you're now asking me,
but I'd suggest first looking at BBT (1997), HS (1965/1975),
and O (1980).]
In case anyone's interested, below are some of the references
I had on hand when I was working on that survey of Lusin's
theorem (which my sci.math post 7-8 months later was a
summary of), a survey that remains to be completed,
by the way.
Daniel C. Biles and John S. Spraker, "Continuous almost
everywhere", International Journal of Mathematical
Education in Science and Technology 30 (1999), 287-289.
Henry Blumberg, "New properties of all real functions",
Transactions of the American Mathematical Society 24
(1922), 113-128.
Jack B. Brown, "Metric spaces in which a strengthened
form of Blumberg's theorem holds", Fundamenta Mathematicae
71 (1971), 243-253.
Jack B. Brown, "Lusin density and Ceder's differentiable
restrictions of arbitrary real functions", Fundamenta
Mathematicae 84 (1974), 35-45.
Jack B. Brown, Continuous-, derivative-, and differentiable-
restrictions of measurable functions", Fundamenta
Mathematicae 141 (1992), 85-95.
Jack B. Brown, "Restriction theorems in real analysis",
Real Analysis Exchange 20 (1994-95), 510-526.
Jack B. Brown, "C^1-intersection variant of Blumberg's
theorem", Tatra Mountains Mathematical Publications 14
(1998), 127-136.
Jack B. Brown and Karel Prikry, "Variations on Lusin's
theorem", Transactions of the American Mathematical
Society 302 (1987), 77-86.
Leon Warren Cohen, "A new proof of Lusin's theorem",
Fundamenta Mathematicae 9 (1927), 122-123.
Robert Bruce Crofoot, "Continuity on a set", College
Mathematics Journal 26 (1995), 29-30.
Eric K. van Douwen, "Nonextendible functions characterize
dense G_delta's in metrizable spaces", Topology and its
Applications 51 (1993), 187-190.
Steven N. Evans and James William Pitman, "Does every Borel
function have a somewhere continuous modification?", Real
Analysis Exchange 18 (1992-93), 276-280.
Marcus B. Feldman, "A proof of Lusin's theorem", The American
Mathematical Monthly 88 (1981), 191-192.
Casper Goffman, "Proof of a theorem of Saks and Sierpinski",
Bulletin of the American Mathematical Association 54 (1948),
950-952.
Casper Goffman, "Lusin's theorem", Real Analysis Exchange
16 (1990-91), 382-392.
Casper Goffman and Fon Che Liu, "Lusin type theorems for
functions of bounded variation", Real Analysis Exchange
5 (1979), 261-266.
William J. Gorman, "Lebesgue equivalence to functions of
the first Baire class", Proceedings of the American
Mathematical Society 17 (1966), 831-834.
Ondrej F. K. Kalenda and Jairi Spurny, "Extending Baire-one
functions on topological spaces", manuscript, May 2004, 25 pages.
Vladimir G. Kanove [Kanovej], "The development of the descriptive
theory of sets under the influence of the work of Luzin",
Russian Mathematical Surveys 40 #3 (1985), 135-180.
Miklos Laczkovich, "Baire 1 functions", Real Analysis Exchange
9 (1983-84), 15-28.
Miklos Laczkovich, "Differentiable restrictions of continuous
functions", Acta Mathematica Hungarica 44 (1984), 355-360.
Denny H. Leung and Wee-Kee Tang, "Functions of Baire class one",
Fundamenta Mathematicae 179 (2003), 225-247.
Norman Levine, "A note on functions continuous almost everywhere",
The American Mathematical Monthly 66 (1959), 791-792.
Edgar Raymond Lorch, "Continuity and Baire functions", The American
Mathematical Monthly 78 (1971), 748-762.
Edward Marczewski [Szpilrajn] and Tadeusz Traczyk, "On developable
sets and almost-limit points", Colloquium Mathematicum 8 (1961),
55-66.
R. Daniel Mauldin, "\sigma-ideals and related Baire systems",
Fundamenta Mathematicae 71 (1992), 171-177.
R. Daniel Mauldin, "The Baire order of the functions continuous
almost everywhere", Proceedings of the American Mathematical
Society 41 (1973), 535-540.
Michal Morayne, "Algebras of Borel measurable functions",
Fundamenta Mathematicae 141 (1992), 229-242.
Arthur H. Stone, "Lusin's theorem", Atti del Seminario Matematico
e Fisico dell'Universita di Modena 44 (1996), 351-357.
Ivo Vrkoc, "Remark about the relation between measurable and
continuous functions", Casopis Pro Pestovani Matematiky
[= Mathematica Bohemica] 96 (1971), 225-228.
Albert Wilansky, "Two examples in real variables", The American
Mathematical Monthly 60 (1953), 317. [A correction is given in
Amer. Math. Monthly 60 (1953), 546.]
Dave L. Renfro
jennifer
Thanks dave, appreciate it.
cbr...@cbrownsystems.com
> Dave L. Renfro wrote:
> > Robert Israel wrote:
> >
> > >> Of course she still won't know an example of a
> > >> non-measurable function. Nor will she ever, for
> > >> some interpretations of "know an example".
> >
> > jennifer wrote:
> >
> > > sexist pig
> >
> > He's talking about set-theoretic definability issues relating
> > to things like the axiom of choice.
<snip>
> > Dave L. Renfro
>
> what if e= 0, as the assumption that i am deleting?
Just a point of (N)etiquette:
We all make mistakes; and it's understandable, given the low bandwidth
of Usenet, that brevity is sometimes mistaken for sarcasm or worse. But
if you want future expert help on sci.math, you should consider
apologizing for insulting Robert Israel with the label "sexist pig".
He's not a sexist pig on sci.math; as can be seen by reviewing his
posting record.
Cheers - Chas
Dave L. Renfro
> Thanks dave, appreciate it.
I just wrote a reply to another post of yours that
I've already replied to. What happened is that
one of your posts appeared twice, and I mistakenly
thought the second appearance was in reply to
the post I made yesterday (instead of in reply to
a post I made two days ago). Since I've already
written it, I may as well post it, as it might
be of use to you.
For e = 0, consider the characteristic function of a
set that is nowhere dense with positive measure.
Proof: Let f be the characteristic function of such a set
and assume g = f a.e. Then there exists a point p such
that f(p) = g(p) = 1 (why?). In every neighborhood of p,
the points where f = 0 contains an open interval (why?),
so in every neighborhood of p there exists a point where
g takes on the value of 0 (why?). The last two sentences
imply that g cannot be continuous at the point p. Hence,
there does not exist a continuous function g such that
g = f a.e.
For measurability of f, look in the books I cited or in
some of the papers I cited (yes, I know, you may have
to actually have to visit a university library, but I've
done the hard part for you by telling you where to look)
for a proof that Lusin's property implies measurability
of the function. I don't remember how the proof goes, or
even how easy it is, although I do know that it is not
too difficult to present in an average graduate real
analysis class. Another book you might want to look at
if you make it to a library is
A. B. Kharazishvili, "Nonmeasurable Sets and Functions",
North-Holland Mathematics Studies #195, Elsevier, 2004.
Also, the following book by the same guy (whose last name
apparently has more than one English version) has some
information about Lusin's theorem in Chapter 7. Use
"Search in this book" = 'Luzin', begin with p. 149,
and when you get to your 2-page page advance limit,
switch to "Search in this book" = 'Blumberg'.
"Strange Functions in Real Analysis"
by Aleksandr B. Charazisvili
http://books.google.com/books?vid=ISBN1584885823
Dave L. Renfro
The World Wide Wade
<1163697546.6...@e3g2000cwe.googlegroups.com>,
"Dave L. Renfro" <renf...@cmich.edu> wrote:
> jennifer wrote:
>
> > Thanks dave, appreciate it.
>
> I just wrote a reply to another post of yours that
> I've already replied to. What happened is that
> one of your posts appeared twice, and I mistakenly
> thought the second appearance was in reply to
> the post I made yesterday (instead of in reply to
> a post I made two days ago). Since I've already
> written it, I may as well post it, as it might
> be of use to you.
>
> For e = 0, consider the characteristic function of a
> set that is nowhere dense with positive measure.
Or on [0,1/2] let f be the characteristic function of [0,1/2].
Suppose g is continuous and g = f a.e. In any neighborhood of
1/2, g takes on the values 0 and 1, contradiction
Dave L. Renfro
>> For e = 0, consider the characteristic function of a
>> set that is nowhere dense with positive measure.
The World Wide Wade wrote:
> Or on [0,1/2] let f be the characteristic function of [0,1/2].
> Suppose g is continuous and g = f a.e. In any neighborhood of
> 1/2, g takes on the values 0 and 1, contradiction
Ah, much simpler! I seem to recall a homework problem I
assigned in a real analysis course back in Fall 2001 where
I asked students to give a careful proof that the characteristic
function of an interval (not equal to R) is not a.e. equal to
some continuous function (and yet it has at most two points of
discontinuity), so I'm especially curious about why I would
have used fat Cantor sets for such an example. There might
be a reason, buried in my notes on the Lusin's theorem
material that I wrote 2 years ago (and looked back over and
posted a summary of 1.5 years ago and haven't looked at since),
for using this example, a reason I overlooked when I wrote that
post, I don't know. But I'll jot this example on a post-it note
and stick it with my Lusin's theorem folders for consideration
when I return to that topic.
Dave L. Renfro
The World Wide Wade
<1163774882.0...@b28g2000cwb.googlegroups.com>,
"Dave L. Renfro" <renf...@cmich.edu> wrote:
A countable union of fat Cantor sets E, if designed properly,
will give rise to measurable function, namely X_E, such that for
every interval I and every continuous function g on I, m{x in I :
X_E =! g} > 0.
bobm2005
no intervals? If so, it makes some of my proofs damn awkward!
Arturo Magidin
bobm2005 <bobm...@postmaster.co.uk> wrote:
>Can a set (let's say on the real line) of non-zero measure contain
>no intervals? If so, it makes some of my proofs damn awkward!
Yes. Let X be any finite interval, and let A= X\Q, where Q is the set
of all rationals. Since the rationals are of measure zero,
m(A)=m(X)>0, but A contains no intervals.
By the by: don't top-post.
http://www.xs4all.nl/~hanb/documents/quotingguide.html
http://www.caliburn.nl/topposting.html
http://www.html-faq.com/etiquette/?toppost
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
The World Wide Wade
<1165334410....@80g2000cwy.googlegroups.com>,
"bobm2005" <bobm...@postmaster.co.uk> wrote:
> Can a set (let's say on the real line) of non-zero measure contain
> no intervals? If so, it makes some of my proofs damn awkward!
Certainly. If U is an open set of small measure containing the
rationals, then E = R \ U is such a set. For all nontrivial
intervals I, m(I intersect E) < m(I).
bobm2005
> In article <1165334410....@80g2000cwy.googlegroups.com>,
> bobm2005 <bobm...@postmaster.co.uk> wrote:
> >Can a set (let's say on the real line) of non-zero measure contain
> >no intervals? If so, it makes some of my proofs damn awkward!
>
> Yes. Let X be any finite interval, and let A= X\Q, where Q is the set
> of all rationals. Since the rationals are of measure zero,
> m(A)=m(X)>0, but A contains no intervals.
Sounds good. Let's say we take S = all irrationals in [-1,1].
But hang on - are the rationals in [-1,1] a measurable set?
Bear in mind the definition ultimately comes down to step functions...
A N Niel
bobm2005 <bobm...@postmaster.co.uk> wrote:
> Sounds good. Let's say we take S = all irrationals in [-1,1].
> But hang on - are the rationals in [-1,1] a measurable set?
> Bear in mind the definition ultimately comes down to step functions...
Have you proved that the collection of measurable sets is
closed under certain countable operations?
Dave Seaman
Every countable set is measurable.
--
Dave Seaman
U.S. Court of Appeals to review three issues
concerning case of Mumia Abu-Jamal.
<http://www.mumia2000.org/>
Arturo Magidin
>> In article <1165334410....@80g2000cwy.googlegroups.com>,
>> bobm2005 <bobm...@postmaster.co.uk> wrote:
>> >Can a set (let's say on the real line) of non-zero measure contain
>> >no intervals? If so, it makes some of my proofs damn awkward!
>>
>> Yes. Let X be any finite interval, and let A= X\Q, where Q is the set
>> of all rationals. Since the rationals are of measure zero,
>> m(A)=m(X)>0, but A contains no intervals.
>
>Sounds good. Let's say we take S = all irrationals in [-1,1].
>But hang on - are the rationals in [-1,1] a measurable set?
Yes.
>Bear in mind the definition ultimately comes down to step functions...
It does? Here I thought measurability of subsets came down to
intervals, and that it was measurability of ->functions<- which was
related to step functions. In any case:
Every open interval is measurable. Therefore, so is the complement of
a single point, since it of the form (-oo,a)U(a,oo), which is
measurable (being the union of two measurable sets). Therefore, every
singleton {a} is measurable, being the complement of a measurable
set. Since measurability is closed under countable unions, every
countable subset of R is necessarily measurable. Therefore, every
subset of the rationals is measurable (and of measure 0, since the
measure is sigma-additive). Therefore, the complement of any subset of
rationals is measurable, and in particular, the set of all irrationals
is measurable.
Since any interval (open, closed, semi-open, semiclosed, whatever) is
measurable, and the intersection of two measurable sets is measurable,
it follows that "all irrationals in [-1,1]", the intersection of the
measurable set [-1,1] and the measurable set "all irrationals", is
measurable.
Can I let go now, or should I still hang on?
Michael Press
mag...@math.berkeley.edu (Arturo Magidin) wrote:
I appreciate that you have stayed this long, and shall
not hold you. Thanks.
--
Michael Press
bobm2005
> >Sounds good. Let's say we take S = all irrationals in [-1,1].
> >But hang on - are the rationals in [-1,1] a measurable set?
>
> Yes.
>
> >Bear in mind the definition ultimately comes down to step functions...
>
> It does? Here I thought measurability of subsets came down to
> intervals, and that it was measurability of ->functions<- which was
> related to step functions. In any case:
>
> Every open interval is measurable. Therefore, so is the complement of
> a single point, since it of the form (-oo,a)U(a,oo), which is
> measurable (being the union of two measurable sets). Therefore, every
> singleton {a} is measurable, being the complement of a measurable
> set. Since measurability is closed under countable unions, every
> countable subset of R is necessarily measurable. Therefore, every
> subset of the rationals is measurable (and of measure 0, since the
> measure is sigma-additive). Therefore, the complement of any subset of
> rationals is measurable, and in particular, the set of all irrationals
> is measurable.
>
> Since any interval (open, closed, semi-open, semiclosed, whatever) is
> measurable, and the intersection of two measurable sets is measurable,
> it follows that "all irrationals in [-1,1]", the intersection of the
> measurable set [-1,1] and the measurable set "all irrationals", is
> measurable.
>
> Can I let go now, or should I still hang on?
>
[-1,1]
is zero for all v in C^\infty_0[-1,1]. Can you prove f is zero a.e.? I
have
heard this referred to as the DuBois-Reymond lemma, but can find no
reference to it. It relates in an obvious way to the previous
questions.
Ideally I want f to be L^2.
The World Wide Wade
<1165501935.3...@j72g2000cwa.googlegroups.com>,
"bobm2005" <bobm...@postmaster.co.uk> wrote:
Yes I can.