TITLE: Do probability measures have to be the same if they agree on a generator of Borel $\sigma$–algebra $\mathcal{B}(\mathbb{R})$?
QUESTION [6 upvotes]: 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.

REPLY [8 votes]: 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.

REPLY [6 votes]: 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 $(\mathbb R,\mathcal B(\mathbb R))$.