S.O.S. Mathematics CyberBoard

Your Resource for mathematics help on the web!
It is currently Sat, 25 May 2013 00:27:57 UTC

All times are UTC [ DST ]




Post new topic Reply to topic  [ 4 posts ] 
Author Message
 Post subject: Measurability
PostPosted: Mon, 23 Mar 2009 10:52:54 UTC 
Offline
Senior Member
User avatar

Joined: Sat, 15 Dec 2007 15:57:40 UTC
Posts: 67
Location: Kyoto
Let (X,A, \mu) be a measurable space with \mu(X) < 1, and let f : (X,A) \rightarrow (R, B) be
an integrable positive function (B is Borel-set), i.e. satisfying
\int_X fd\mu < \infty
Show that the integral can be calculated as the limit
\int_X fd\mu = \lim_{n\rightarrow\infty} \sum_{k=0}^{\infty}z_{n,k} \mu\{x\in X : y_{n,k} \leq f(x) < y_{n,k+1}\}
where for every n\in N
0 < y_{n,0} < y_{n,1} < y_{n,2} < \ldots, \lim_{k\rightarrow\infty}y_{n,k} = \infty
z_{n,k} is any number satisfying y_{n,k} \leq z_{n,k} \leq y_{n,k+1} and
\lim_{n\rightarrow\infty}\sup_{k\geq0}\left(y_{n,k+1}-y_{n,k}\right)

I actually understand the question, I just have no idea how to do it, how to write it down or work it out.


Top
 Profile  
 
 Post subject: Re: Measurability
PostPosted: Mon, 23 Mar 2009 14:41:04 UTC 
Offline
Moderator
User avatar

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6009
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
Sarah wrote:
Let (X,A, \mu) be a measurable space with \mu(X) < 1, and let f : (X,A) \rightarrow (R, B) be
an integrable positive function (B is Borel-set), i.e. satisfying
\int_X fd\mu < \infty
Show that the integral can be calculated as the limit
\int_X fd\mu = \lim_{n\rightarrow\infty} \sum_{k=0}^{\infty}z_{n,k} \mu\{x\in X : y_{n,k} \leq f(x) < y_{n,k+1}\}
where for every n\in N
0 < y_{n,0} < y_{n,1} < y_{n,2} < \ldots, \lim_{k\rightarrow\infty}y_{n,k} = \infty
z_{n,k} is any number satisfying y_{n,k} \leq z_{n,k} \leq y_{n,k+1} and
\lim_{n\rightarrow\infty}\sup_{k\geq0}\left(y_{n,k+1}-y_{n,k}\right)

I actually understand the question, I just have no idea how to do it, how to write it down or work it out.


I think your last condition should be
\displaystyle
\lim_{n\to\infty}\sup_{k\geq 0}\left(y_{n,k+1}-y_{n,k}\right)=0.
If f is a simple function the result holds. So approximate f by simple functions ...


Top
 Profile  
 
 Post subject: Re: Measurability
PostPosted: Mon, 23 Mar 2009 14:41:44 UTC 
Offline
Member of the 'S.O.S. Math' Hall of Fame
User avatar

Joined: Sat, 7 Jan 2006 18:29:24 UTC
Posts: 1401
Location: Leeds, UK
Sarah wrote:
Let (X,A, \mu) be a measurable space with \mu(X) < 1, and let f : (X,A) \rightarrow (R, B) be
an integrable positive function (B is Borel-set), i.e. satisfying
\int_X fd\mu < \infty
Show that the integral can be calculated as the limit
\int_X fd\mu = \lim_{n\rightarrow\infty} \sum_{k=0}^{\infty}z_{n,k} \mu\{x\in X : y_{n,k} \leq f(x) < y_{n,k+1}\}
where for every n\in N
0 < y_{n,0} < y_{n,1} < y_{n,2} < \ldots, \lim_{k\rightarrow\infty}y_{n,k} = \infty
z_{n,k} is any number satisfying y_{n,k} \leq z_{n,k} \leq y_{n,k+1} and
\lim_{n\rightarrow\infty}\sup_{k\geq0}\left(y_{n,k+1}-y_{n,k}\right)

I actually understand the question, I just have no idea how to do it, how to write it down or work it out.

Let g_n(x) = \sum_{k=0}^{\infty}z_{n,k}\chi_{n,k}(x), where \chi_{n,k}(x) is the function that takes the value 1 on the set \{x\in X : y_{n,k} \leq f(x) < y_{n,k+1}\} and is zero elsewhere. Notice that g_k is integrable, with integral \sum_{k=0}^{\infty}z_{n,k} \mu\{x\in X : y_{n,k} \leq f(x) < y_{n,k+1}\}, and that |g_k(x) - f(x)|\leqslant \sup_{k\geq0}\left(y_{n,k+1}-y_{n,k}\right) for all x in X. Thus \Bigl|\int_Xf\,d\mu - \int_Xg_n\,d\mu\Bigr|\leqslant \mu(X)\sup_{k\geq0}\left(y_{n,k+1}-y_{n,k}\right).


Top
 Profile  
 
 Post subject:
PostPosted: Mon, 23 Mar 2009 21:36:03 UTC 
Offline
Senior Member
User avatar

Joined: Sat, 15 Dec 2007 15:57:40 UTC
Posts: 67
Location: Kyoto
Of course the last condition is incorrect, thanks for noticing!
The help looks great, I going to write it down completely...


Top
 Profile  
 
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 4 posts ] 

All times are UTC [ DST ]


Who is online

Users browsing this forum: No registered users


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum

Search for:
Jump to:  
Contact Us | S.O.S. Mathematics Homepage
Privacy Statement | Search the "old" CyberBoard

users online during the last hour
Powered by phpBB © 2001, 2005-2011 phpBB Group.
Copyright © 1999-2013 MathMedics, LLC. All rights reserved.
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA