## Re: [harmonic] L^/infinity functions

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• Quoting gnavada : It s false.
Message 1 of 7 , Feb 13, 2007
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It's false.

> Hello,
> I need a proof of the following fact: L^/infinity functions in R can
> be made continuous by changing the values of functions on a set of
> measure zero. Even a reference to the proof will help.
> thanks
> gowri
>
>
• Hai, I think that is not true in general. Particular you can visualise this by taking a step function on a bounded interval. bye.... wilson@cems.uvm.edu wrote:
Message 2 of 7 , Feb 13, 2007
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Hai,
I think that is not true in general. Particular you can visualise this by taking a step function on a bounded interval.
bye....

wilson@... wrote:

It's false.

> Hello,
> I need a proof of the following fact: L^/infinity functions in R can
> be made continuous by changing the values of functions on a set of
> measure zero. Even a reference to the proof will help.
> thanks
> gowri
>
>

D.VENKU NAIDU,
CAUVERY HOSTEL,
ROOM NO:2045,
CHENNAI-36

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• Hi Thanks. I got similar mails from others saying it is not true. I havnt checked a counterexample yet, but I ll do it. In page 192 of Rudin s Real and
Message 3 of 7 , Feb 13, 2007
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Hi
Thanks. I got similar mails from others saying it is
not true. I havnt checked a counterexample yet, but
I'll do it.
In page 192 of Rudin's "Real and Complex Analysis", he
uses a /beta which is in L^/infinity and then he says
(after equation (5)) that by changing /beta(y)on a set
of measure 0 we may assume it is continuous. If you
have some time to spare can u pl. go through this pf
and explain that step?
Thank you very much
regards,
gowri

--- venku naidu <venku_uoh@...> wrote:

> Hai,
> I think that is not true in general. Particular
> you can visualise this by taking a step function on
> a bounded interval.
> bye....
>
> wilson@... wrote:
>
> It's false.
>
> > Hello,
> > I need a proof of the following fact: L^/infinity
> functions in R can
> > be made continuous by changing the values of
> functions on a set of
> > measure zero. Even a reference to the proof will
> help.
> > thanks
> > gowri
> >
> >
>
>
>
>
>
>
> D.VENKU NAIDU,
> CAUVERY HOSTEL,
> ROOM NO:2045,
> CHENNAI-36
>
>
> ---------------------------------
> Now you can scan emails quickly with a reading
> pane. Get the new Yahoo! Mail.

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• Hi Gowri, Look at equation (5) on pg. 192. We assumed, that phi(f) is not zero and that beta(y) is in L^ infinity. On the other hand, since the map f to f_y
Message 4 of 7 , Feb 14, 2007
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Hi Gowri,

Look at equation (5) on pg. 192. We assumed, that \phi(f) is not zero and that \beta(y) is in L^\infinity. On the other hand, since the map f \to f_y (translation by 'y') is continuous and the function \phi is a continuous functional we have a continuous function on the right hand side. Eq. (5) says that they are the same a.e. Change the value of  \beta(y) to (1/\phi(f)) \phi(f_y) for all 'y' at which eq. (5) doesn't hold. You end up with a new \beta function that agrees with the old one a.e. and agrees with (1/\phi(f)) \phi(f_y) everywhere. This latest function is continuous.

Note, that the situation is special since the measure '\beta dm' comes from the Riesz representation theorem and definitely not arbitrary!

Bela

Hi
Thanks. I got similar mails from others saying it is
not true. I havnt checked a counterexample yet, but
I'll do it.
In page 192 of Rudin's "Real and Complex Analysis", he
uses a /beta which is in L^/infinity and then he says
(after equation (5)) that by changing /beta(y)on a set
of measure 0 we may assume it is continuous. If you
have some time to spare can u pl. go through this pf
and explain that step?
Thank you very much
regards,
gowri

--- venku naidu <venku_uoh@...> wrote:

> Hai,
> I think that is not true in general. Particular
> you can visualise this by taking a step function on
> a bounded interval.
> bye....
>
> wilson@... wrote:
>
> It's false.
>
> > Hello,
> > I need a proof of the following fact: L^/infinity
> functions in R can
> > be made continuous by changing the values of
> functions on a set of
> > measure zero. Even a reference to the proof will
> help.
> > thanks
> > gowri
> >
> >
>
>
>
>
>
>
> D.VENKU NAIDU,
> CAUVERY HOSTEL,
> ROOM NO:2045,
> CHENNAI-36
>
>
> ---------------------------------
> Now you can scan emails quickly with a reading
> pane. Get the new Yahoo! Mail.

__________________________________________________________
Be a PS3 game guru.
Get your game face on with the latest PS3 news and previews at Yahoo! Games.
http://videogames.yahoo.com/platform?platform=120121

--
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Albert Schweitzer
• Hi Gowri; I think the best you can do in approximating such function is Lusin s Theorem, which is sometimes very useful. Anyways, here I provide a counter
Message 5 of 7 , Mar 3, 2007
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Hi Gowri;

I think the best you can do in approximating such function is Lusin's
Theorem, which is sometimes very useful. Anyways, here I provide a
counter example to your result (which I hope is right).

There exists a sequence {En} of disjoint measurable sets on R such
that
(i) the measure of En is less than 2^n
(ii)there exists a countable basis {In} of the topology of R then En
intersects In in a set of positive measure for every n
(note that you use this construction to solve exercise 8 of Rudin
Chapter 2)

Now consider the function f=sigma{(1/2)^n * (Characteristic function
of En)/ (measure of En)}. Then since the En are disjoint f converges
everywhere. It is easy to see that the L1 norm of f is finite, but its
integral on any interval of R is infinite. This last fact proves that
even if you change f on a set of measure 0, it will remain
discontinuous.

But you tell me: "f may not be in L^infinity". Then take the function
g=exp{-f} then g is bounded and is continuous if and only if f is
continuous (here we needed the fact that f is never infinite, which
explains my choice of this particular f rather than more easier
examples satisfying almost the same properties as f).

g is in L^infinity (bounded). If I can change g on a set of measure 0
to make it continuous then the same is true for f, which is a