Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
5  
Yes, $\delta_0$ for example. (Dirac at $0$). I guess you want additional conditions. –  Davide Giraudo Sep 22 '12 at 11:43
1  
As @DavideGiraudo said, you probably want more conditions. I am still looking for a natural measure (on $\mathcal{P}([0,1])$). Though I think I now know such an example. Let me know if you're interested in that. –  Quinn Culver Sep 22 '12 at 12:42
1  
To read about the space $\mathcal{P}(\mathcal{P}(\mathbb{R}))$, check out Billingsley's Convergence of Probability Measures and Parthasarathy's Probability Measures on Metric Spaces. –  Quinn Culver Sep 22 '12 at 12:45
    
If you drop the axiom of choice then you can have the Lebesgue measure... –  Asaf Karagila Sep 22 '12 at 13:19
    
@Quinn I think he means power set by ${\cal P}$, not the space of probability measures. –  Byron Schmuland Sep 22 '12 at 14:49
show 2 more comments

1 Answer

up vote 5 down vote accepted

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 $x\in I$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.