- Quoting gnavada <gnavada@...>:

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....
wrote:*wilson@...*Quoting gnavada <gnavada@yahoo. com>:

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,

IIT,MADRAS,

CHENNAI-36

Now you can scan emails quickly with a reading pane. Get the new Yahoo! Mail. - 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:

> Quoting gnavada <gnavada@...>:

>

> 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,

> IIT,MADRAS,

> 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 - 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!

BelaOn 2/14/07,**Gowri Navada**<gnavada@...> wrote: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:

> Quoting gnavada <gnavada@...>:

>

> 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,

> IIT,MADRAS,

> 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

--

"Az ember egy okos állat, amelyik úgy viselkedik, mint egy hülye."

"Man is a clever animal who behaves like an imbecile."

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 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

contradiction.

Please check the above counter example and keep me in touch if

something is wrong.

Best Regards;

Zaher Hani