simple question about measurable characteristic function


by sunjin09
Tags: characteristic, function, measurable, simple
sunjin09
sunjin09 is offline
#1
May24-12, 04:13 PM
P: 312
1. The problem statement, all variables and given/known data
Prove that the characteristic function [itex]\chi_A: X\rightarrow R, \chi_A(x)=1,x\in A; \chi_A(x)=0, x\notin A[/itex], where A is a measurable set of the measurable space [itex] (X,\psi) [/itex], is measurable.


2. Relevant equations
a function [itex]f: X->R[/itex] is measurable if for any usual measurable set M of R, [itex]f^{-1}(M)[/itex] is measurable in [itex](X,\psi)[/itex]

3. The attempt at a solution
Obviously [itex]f^{-1}([0,1])=X[/itex], where the universal set X need not be a measurable set in a general measurable space [itex](X,\psi)[/itex], which only requires that the (uncountable) union of all measurable sets is X. But the book explicitly asked to prove for a general measurable space. What am I missing here? Thank you.
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Citan Uzuki
Citan Uzuki is offline
#2
May24-12, 04:47 PM
PF Gold
P: 274
Quote Quote by sunjin09 View Post
3. The attempt at a solution
Obviously [itex]f^{-1}([0,1])=X[/itex], where the universal set X need not be a measurable set in a general measurable space [itex](X,\psi)[/itex], which only requires that the (uncountable) union of all measurable sets is X. But the book explicitly asked to prove for a general measurable space. What am I missing here? Thank you.
[itex]X=A\cup A^{c}[/itex]
sunjin09
sunjin09 is offline
#3
May24-12, 06:40 PM
P: 312
But [itex]A^c[/itex] need not be measurable in a general measurable space, which is not necessarily a Borel field, only a [itex]\sigma[/itex]-ring whose union is X. Am I completely wrong?

Vargo
Vargo is offline
#4
May24-12, 08:34 PM
P: 350

simple question about measurable characteristic function


I've heard of sigma fields and sigma algebras before, but never sigma rings... Is that a variation in which measurability is not closed under complements? I would double check that in your book.
Citan Uzuki
Citan Uzuki is offline
#5
May24-12, 09:20 PM
PF Gold
P: 274
Every book I've seen defines a measurable space as a set equipped with a Σ-algebra. What book are you using?
sunjin09
sunjin09 is offline
#6
May25-12, 02:06 PM
P: 312
This is the book I use
http://books.google.com/books/about/...d=9PXuAAAAMAAJ

The definition of general measurable space in this book
Definition 7.1(3). Let X be a (universal) set and let psi be a sigma-ring on X which has the property that X is a (not necessarily countable) union of sets taken from the collection psi. Then the ordered pair (X, psi) is called a measurable space. The members of psi are referred to as the measurable sets of X.

A example in the book is let X be an uncountable set and psi be all countable subset of X.

This is a wiki page on sigma-ring:
http://en.wikipedia.org/wiki/Sigma-ring
In the last paragraph:
"σ-rings can be used instead of σ-fields in the development of measure and integration theory, if one does not wish to require that the universal set be measurable."

In the same book there is this problem:
Let (X, psi) be a general measurable space. Show that the characteristic function of [itex]A\subset X[/itex] is measurable iff A is measurable. ( Remark: In the text we gave the proof for a Borel space only. )

P.S. I seem to have figured it out. In the book, for a general measurable space, a measurable function f is defined in such a way that the set on which f is 0 is eliminated. Since [itex]\chi_A(A^c)=0, A^c[/itex] need not be measurable. Thank you both!


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