If any two sets on a $\pi$-system with certain properties have equal probability, then is the same true on the generated $\sigma$-algebra? Let $(\Omega,\Sigma)$ be a measurable space and let $X:\Omega\to\mathbb R$ be a random variable that is measurable $\Sigma/\mathscr B$. Here, $\mathscr B$ denotes the Borel $\sigma$-algebra on $\mathbb R$. Suppose furthermore that


*

*$\mathscr P$ is a $\pi$-system on $\mathbb R$ (closed under finite intersections) such that $\mathscr P$ generates the Borel $\sigma$-algebra $\mathscr B$;

*$\nu$ is a Borel probability measure on $\mathbb R$;

*if $H_1,H_2,\in\mathscr P$ are such that $X^{-1}(H_1)=X^{-1}(H_2)$, then $\nu(H_1)=\nu(H_2)$.
NOTE: $\nu$ is not assumed to be the probability distribution of the random variable $X$.

Conjecture: If $H_1,H_2,\in\mathscr B$ are such that $X^{-1}(H_1)=X^{-1}(H_2)$, then $\nu(H_1)=\nu(H_2)$.

That is, I want to extend the probabilistic indistinguishability of two sets such that the preimages of $X$ coincide on the $\pi$-system to the generated $\sigma$-algebra. I tried using Dynkin’s theorem, but I hit a dead end. Any hints would be greatly appreciated.
 A: The statement is false. Let $(\Omega,\Sigma)$ be an arbitrary measurable space and let $X(\omega)=0$, $\omega\in\Omega$, be constant.
Let $\mathscr P$ be the collection of those Borel sets that do not contain the point $1$. This is clearly a $\pi$-system (the intersection of a finite number of Borel sets missing the point $1$ is a Borel set not containing $1$). Furthermore, $\mathscr P$ generates $\mathscr B$. To see this, suppose that $B\in\mathscr B$. If $1\notin B$, then $B\in\mathscr P$ by definition. If $1\in B$, then $B\setminus\{1\}$ is in $\mathscr P$ and the singleton $\{1\}$ is in the $\sigma$-algebra $\sigma(\mathscr P)$ generated by $\mathscr P$, given that $$(-\infty,1)\cup(1,\infty)\in\mathscr P,$$ and $\{1\}$ is the complement of the latter set. Therefore, $ B= B\setminus\{1\}\cup\{1\}\in\sigma(\mathscr P)$. It follows that $\mathscr B\subseteq\sigma(\mathscr P)$ and the other direction of inclusion is obvious.
Now let $\nu$ be the unit mass at the point $1$. If $H_1$ and $H_2$ are in $\mathscr P$, then $\nu(H_1)=\nu(H_2)=0$ always holds, given that neither set contains $1$. However, if $H_1=[0,1]$ and $H_2=\{0\}$, then $$X^{-1}(H_1)=X^{-1}(H_2)=\Omega,$$ yet $$\nu(H_1)=1\neq 0=\nu(H_2).$$
