Simple question about measurable characteristic function

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Homework Help Overview

The discussion revolves around the measurability of the characteristic function \(\chi_A\) defined on a measurable space \((X, \psi)\). The original poster seeks to prove that this function is measurable, given that \(A\) is a measurable set within the space.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the definition of measurability in the context of general measurable spaces, questioning the implications of the universal set \(X\) not being measurable. There is a focus on the characteristics of \(\sigma\)-rings versus \(\sigma\)-algebras and their impact on the measurability of sets.

Discussion Status

The discussion is ongoing, with participants examining different interpretations of measurability and the definitions provided in various texts. Some guidance has been offered regarding the implications of \(\sigma\)-rings, and the original poster appears to be synthesizing information to clarify their understanding.

Contextual Notes

There is a specific mention of the definitions from a particular textbook, which may differ from standard definitions in other literature. The original poster notes that the characteristic function's behavior regarding the complement of \(A\) raises questions about the necessity of measurability in that context.

sunjin09
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Homework Statement


Prove that the characteristic function \chi_A: X\rightarrow R, \chi_A(x)=1,x\in A; \chi_A(x)=0, x\notin A, where A is a measurable set of the measurable space (X,\psi), is measurable.


Homework Equations


a function f: X->R is measurable if for any usual measurable set M of R, f^{-1}(M) is measurable in (X,\psi)

The Attempt at a Solution


Obviously f^{-1}([0,1])=X, where the universal set X need not be a measurable set in a general measurable space (X,\psi), 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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sunjin09 said:

The Attempt at a Solution


Obviously f^{-1}([0,1])=X, where the universal set X need not be a measurable set in a general measurable space (X,\psi), 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.

X=A\cup A^{c}
 
But A^c need not be measurable in a general measurable space, which is not necessarily a Borel field, only a \sigma-ring whose union is X. Am I completely wrong?
 
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.
 
Every book I've seen defines a measurable space as a set equipped with a Σ-algebra. What book are you using?
 
This is the book I use
http://books.google.com/books/about/Some_modern_mathematics_for_physicists_a.html?id=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 A\subset X 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 \chi_A(A^c)=0, A^c need not be measurable. Thank you both!
 
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