Is there a measure for every Borel σ-algebra?

[Fredrik]

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?

 

[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)

 

[lavinia]

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

 

[micromass]

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

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.

 

[Fredrik]

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.