# Simple question about measurable characteristic function

1. May 24, 2012

### sunjin09

1. The problem statement, all variables and given/known data
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.

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

3. 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.

2. May 24, 2012

### Citan Uzuki

$X=A\cup A^{c}$

3. May 24, 2012

### sunjin09

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?

4. May 24, 2012

### Vargo

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.

5. May 24, 2012

### Citan Uzuki

Every book I've seen defines a measurable space as a set equipped with a Σ-algebra. What book are you using?

6. May 25, 2012

### sunjin09

This is the book I use

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!

Last edited: May 25, 2012