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.

Suppose $\mathcal{K}\subset 2^\mathbb{R}$ is such that $\sigma(\mathcal{K})=\mathcal{B}(\mathbb{R})$ and let $\mu$ and $\nu$ be measures which agree on $\mathcal{K}$, i.e. $$\mu(A)=\nu(A)$$ for all $A\in\mathcal{K}.$ Do these measures necessarily have to be the same?

I am teaching myself in measure-theoretic probability and for now I can only prove that the answer is yes, assuming $\mathcal{K}$ is a $\pi$-system. However, I do not know what to think when $\mathcal{K}$ is arbitrarily chosen. I would appreciate any hints.

share|improve this question
1  
A closely related question: What is an example of a lambda-system that is not a sigma algebra? This contains what is essentially did's example. –  Nate Eldredge Sep 23 '12 at 15:17

2 Answers 2

up vote 2 down vote accepted

Hint: On the measurable space $(\Omega,2^\Omega)$ with $\Omega=\{1,2,3,4\}$, consider $\mathcal K=\{\{1,2\},\{2,3\}\}$, $\mu=\frac12(\delta_1+\delta_3)$ and $\nu=\frac12(\delta_2+\delta_4)$. Then $\sigma(\mathcal K)=2^\Omega$ and $\mu(A)=\frac12=\nu(A)$ for every $A$ in $\mathcal K$ but $\mu\ne\nu$.

What is left to do: To adapt this counterexample to $(\mathbb R,\mathcal B(\mathbb R))$.

share|improve this answer
    
@Byron Sorry... –  Did Sep 23 '12 at 14:46
    
May the faster typist win! Ha ha... –  Byron Schmuland Sep 23 '12 at 14:48
    
@Byron Yes, this is one not-so-good aspect of the site. (But I left something to do for the OP...) –  Did Sep 23 '12 at 14:50
    
In fact, I have been familiar with examples such as yours or given by Nate Eldredge in his comment and what posed the major difficulty for me was exactly to notice that I simply have to adapt them to the $(\mathbb{R},\mathcal{B}(\mathbb{R})$ case. Thank you for the sharpest hint I could get. Just to make sure I am not mistaken: I simply define $\mathcal{K}=\mathcal{B}(\mathbb{R}\smallsetminus \{1,2,3,4\})\cup \{\{1,2\},\{2,3\},\{3,4\}\}$ which, indeed, generates $\mathcal{B}(\mathbb{R})$. Then the measures described by you agree on $\mathcal{K}$ but are not the same. Did I get it right? –  Kuba Helsztyński Sep 23 '12 at 16:39
    
Yes. Note that $\{3,4\}$ may be omitted from $\mathcal K$ without changing the result. –  Did Sep 23 '12 at 17:06

In this context, it is useful to consider probability measures on a product space, say $\mathbb{R}^2$. Let $\cal K$ be the collection of subsets generated by the projections onto the $x$ and $y$ axes. Then $\cal K$ generates the Borel $\sigma$-field on $\mathbb{R}^2$, and two measures $\mu$ and $\nu$ agree on $\cal K$ exactly when they have the same marginal distributions.

But of course, $\cal K$ is far from being closed under intersections and there are many different measures with the same marginals.

share|improve this answer
    
Indeed. +1. $ $ –  Did Sep 23 '12 at 15:00

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.