# Is there a measure for every Borel σ-algebra?

1. Feb 11, 2012

### Fredrik

Staff Emeritus
The Borel σ-algebra can be defined on every topological space. Does that mean that every topological space can be turned into a measure space?

2. Feb 11, 2012

### micromass

Yes, in a trivial way. Just define $\mu(S)=0$ for all S. Or take the counting measure. Or a point measure.

A better answer is actually Riesz representation theorem. This says that all regular measures on a locally compact Hausdorff space coincide with positive functionals

$$T:C_c(X)\rightarrow \mathbb{R}$$

(with $C_c(X)$ the functions to $\mathbb{R}$ with compact support). Such a functional exist because of the Hahn-Banach theorem.

The same thing can be done with $C_0(X)$ actually. In that case: the measures are the positive functionals, the probability measures are the states and the point measures represent the pure states on X.

In general, the existence of non-trivial measures is not a simple problem and requires set theory (see for example measurable cardinals)

Last edited: Feb 11, 2012
3. Feb 11, 2012

### lavinia

I wonder if the space has a cardinality greater than the continuum whether there are any measures other than the trivial one.

4. Feb 12, 2012

### micromass

The first thing to do is to define the word "the trivial one".

This is quite an interesting theory. In general, this is studied in Boolean algebras and the so-called measure algebras.

5. Feb 12, 2012

### Fredrik

Staff Emeritus
Excellent answer as always. Thanks. I've been meaning to study the Riesz representation theorem. It will be the first thing I do when I'm done with the basics of integration theory.

Last edited: Feb 12, 2012