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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.

Kuba Helsztyński
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2 Answers2

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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.

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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 $(\mathbb R,\mathcal B(\mathbb R))$.

Did
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  • @Byron Sorry... – Did Sep 23 '12 at 14:46
  • May the faster typist win! Ha ha... –  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
  • @did Could you explain, why is $\mathcal{K}$ still a generator of $\mathcal{B}(\mathbb{R})$ then? – Kuba Helsztyński Sep 23 '12 at 17:37
  • Because ${3,4}$ is the complement of the union of $\mathbb R\setminus{1,2,3,4}$ and ${1,2}$. – Did Sep 23 '12 at 18:27