Lipschitz Functions and Riemann Integral
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From: Maury Barbato on 17 May 2010 07:55 Hello, let R be the field of real numbers. Does there exist some Lipschitz function g:[a,b]->[c,d] and some bounded function f:[c,d]-> R wich is Riemann integrable such that f°g is not Riemann integrable? Thank you very much for your attention. My Best Regards, Maury Barbato PS If one only supposes that g is continuous, the answer is yes. Let f be the characteristic function of the ternary Cantor set C, and g a homeomorphism from a "fat" Cantor set onto C. Then f is Riemann integrable on [0,1], but f°g is not Riemann integrable.
From: W^3 on 17 May 2010 17:16 In article <2000510524.174686.1274111768185.JavaMail.root(a)gallium.mathforum.org>, Maury Barbato <mauriziobarbato(a)aruba.it> wrote: > Hello, > let R be the field of real numbers. > Does there exist some Lipschitz function g:[a,b]->[c,d] > and some bounded function f:[c,d]-> R wich is Riemann > integrable such that f°g is not Riemann integrable? > > Thank you very much for your attention. > My Best Regards, > Maury Barbato > > PS If one only supposes that g is continuous, the answer > is yes. Let f be the characteristic function of the > ternary Cantor set C, and g a homeomorphism from a "fat" > Cantor set onto C. Then f is Riemann integrable on [0,1], > but f°g is not Riemann integrable. Let K be a fat Cantor set in [0,1]. Then there exists a C^oo function g : [0,1] -> [0,1] whose zero set is precisely K. Now define f on [0,1] by f(0) = 1, f = 0 elsewhere. Then f is Riemann integrable on [0,1], but f o g is discontinuous at each point of K, a set of positive measure, hence f o g is not Riemann integrable.
From: Ronald Bruck on 21 May 2010 14:42 In article <2000510524.174686.1274111768185.JavaMail.root(a)gallium.mathforum.org>, Maury Barbato <mauriziobarbato(a)aruba.it> wrote: > Hello, > let R be the field of real numbers. > Does there exist some Lipschitz function g:[a,b]->[c,d] > and some bounded function f:[c,d]-> R wich is Riemann > integrable such that f�g is not Riemann integrable? > > Thank you very much for your attention. > My Best Regards, > Maury Barbato > > PS If one only supposes that g is continuous, the answer > is yes. Let f be the characteristic function of the > ternary Cantor set C, and g a homeomorphism from a "fat" > Cantor set onto C. Then f is Riemann integrable on [0,1], > but f�g is not Riemann integrable. I have no idea what this symbol is supposed to be. On my Mac, it looks like an asterisk, but when I save it to file and examine it with od I find it actually reads f "0xc2 0xb0" g whatever that's supposed to be. (Composition?) For not-the-first, and almost certainly not-the-last, time, when you post to Usenet, USE ONLY STRAIGHT ASCII. That will bring all witless OS's, such as Windows, back to reality. (Actually, Windows is perfectly cable of handling Unicode, but nobody's programs, including Microsoft's, seem to utilize it. Perhaps this is Unicode, and my Mac can't handle it. But it's supposed to, out-of-the-box.) --Ron Bruck
From: Frederick Williams on 21 May 2010 14:46 Ronald Bruck wrote: > > In article > <2000510524.174686.1274111768185.JavaMail.root(a)gallium.mathforum.org>, > Maury Barbato <mauriziobarbato(a)aruba.it> wrote: > > > Hello, > > let R be the field of real numbers. > > Does there exist some Lipschitz function g:[a,b]->[c,d] > > and some bounded function f:[c,d]-> R wich is Riemann > > integrable such that f�g is not Riemann integrable? > > > > Thank you very much for your attention. > > My Best Regards, > > Maury Barbato > > > > PS If one only supposes that g is continuous, the answer > > is yes. Let f be the characteristic function of the > > ternary Cantor set C, and g a homeomorphism from a "fat" > > Cantor set onto C. Then f is Riemann integrable on [0,1], > > but f�g is not Riemann integrable. > > I have no idea what this symbol is supposed to be. The circle of function composition. -- I can't go on, I'll go on.
From: Rob Johnson on 21 May 2010 16:55
In article <210520101142328320%bruck(a)math.usc.edu>, Ronald Bruck <bruck(a)math.usc.edu> wrote: >In article ><2000510524.174686.1274111768185.JavaMail.root(a)gallium.mathforum.org>, >Maury Barbato <mauriziobarbato(a)aruba.it> wrote: > >> Hello, >> let R be the field of real numbers. >> Does there exist some Lipschitz function g:[a,b]->[c,d] >> and some bounded function f:[c,d]-> R wich is Riemann >> integrable such that f�g is not Riemann integrable? >> >> Thank you very much for your attention. >> My Best Regards, >> Maury Barbato >> >> PS If one only supposes that g is continuous, the answer >> is yes. Let f be the characteristic function of the >> ternary Cantor set C, and g a homeomorphism from a "fat" >> Cantor set onto C. Then f is Riemann integrable on [0,1], >> but f�g is not Riemann integrable. > >I have no idea what this symbol is supposed to be. On my Mac, it looks >like an asterisk, but when I save it to file and examine it with od I >find it actually reads > > f "0xc2 0xb0" g > >whatever that's supposed to be. (Composition?) > >For not-the-first, and almost certainly not-the-last, time, when you >post to Usenet, USE ONLY STRAIGHT ASCII. That will bring all witless >OS's, such as Windows, back to reality. (Actually, Windows is >perfectly cable of handling Unicode, but nobody's programs, including >Microsoft's, seem to utilize it. Perhaps this is Unicode, and my Mac >can't handle it. But it's supposed to, out-of-the-box.) I use Thunderbird 3.0.4 on OS X 10.5.8, and I see the character as the degree symbol (probably intended as functional composition). The header of Maury's post says >Content-Type: text/plain; charset=UTF-8 >Content-Transfer-Encoding: 8bit and the two byte sequence C2B0 is the UTF-8 encoding for the Unicode character 00B0, which is "DEGREE SIGN". This character doesn't have a standard 7-bit ASCII encoding, however, I had the impression that Thoth 1.8.3 understood UTF-8. What version of OS X are you running? Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font |