1
$\begingroup$

This question has been bugging me for a while? Does there exist a probability measure on the measurable space $\bigl(\mathbb{R},\mathcal{P}(\mathbb{R})\bigr)$. If so, what is it?

  • 0
    @Shahab: You mean perhaps the strongest form of choice for which the Lebesgue measure is a measure on all subsets? We know that this implies that this is consistent with the principle of Dependent Choice, and it immediately negates the following: The weak ultrafilter lemma for $\mathbb N$ (i.e. every ultrafilter on $\mathbb N$ is principal), therefore ultrafilter lemma/Boolean Prime Ideal theorem; Axiom of choice for families of size $\aleph_1$, and therefore $\mathrm{DC}_{\aleph_1}$.2012-09-22

1 Answers 1

6

Reference: You may be interested in the "problem of measure". There is a short treatment of this topic in Appendix C of Real Analysis and Probability by R.M. Dudley.

He proves the following result due to Banach and Kuratowski: Assuming the continuum hypothesis, there is no measure $\mu$ defined on all subsets of $I:=[0,1]$ with $\mu(I)=1$ and $\mu(\{x\})=0$ for all $x\in I$.