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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.

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    A closely related question: [What is an example of a lambda-system that is not a sigma algebra?](http://math.stackexchange.com/questions/8173/what-is-an-example-of-a-lambda-system-that-is-not-a-sigma-algebra) This contains what is essentially did's example.2012-09-23

2 Answers 2

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

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    Because $\{3,4\}$ is the complement of the union of $\mathbb R\setminus\{1,2,3,4\}$ and $\{1,2\}$.2012-09-23
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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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    Indeed. +1. $ $2012-09-23