Let $(X,\Sigma, \mu)$ be a standard Borel space with $\mu$ a finite measure. Consider the space $(X^\mathbb Z, \Sigma_p, \nu)$, where $\Sigma_p$ is the standard $\sigma$ algebra on the product space generated by the projections $\pi_n:X^{\mathbb Z}\to X$, and $\nu$ is the measure on $\Sigma_p$ obtained via the Kolmogorov extension theorem. More specifically $\nu$ is the measure on $\Sigma_p$ which ensures that $(\pi_n)_*\nu =\mu$ for every $n\in \mathbb Z$, which implies that $$\nu\left(\left(\prod_{i=-\infty}^{-m-1} X\right)\times E_{-m}\times\dots\times E_n\times \left(\prod_{k=n+1}^\infty X\right)\right)=\mu(E_{-m})\dots\mu(E_n),$$ for $m,n\in \mathbb N$ and each $E_i\in \Sigma$. This guarantees that the projections are measure preserving. Now let $Y\subset \Sigma_p$, and define $\bar \nu$ on $\Sigma_p\cap Y$ by $\bar\nu(A)=\nu(A)/\nu(Y)$ for all $A\in \Sigma_p\cap Y\subset \Sigma_p$.
My intuition is telling me that if we consider, for any $n$, $\bar\pi_{n}=\pi_n|_Y$ that $\bar \pi_n$ is $\mu,\bar\nu$ measure preserving. However, I'm struggling to prove it. For any $A\in \Sigma$ we have $$\bar\nu(\bar \pi_n^{-1}A)=\frac{\nu(\pi_n^{-1} A\cap Y)}{\nu(Y)},$$ and it feels to me that if the intersection has nonzero measure, then $\nu(\pi_n^{-1}A\cap Y)=\nu(\pi_n^{-1}A)\nu(Y)$, but I cannot prove it. The suggestions of wiser minds would be much appreciated.
This might have to be turned into another question, but if it turns out that such a restriction of the projections is not measure preserving w.r.t $\bar\nu$, is there another natural measure on $\Sigma_p\cap Y$ which makes the restricted coordinate projections measurable?
I believe I have come up with a counterexample. Let $X=[0,1]$, $\mu$ be the Lebesgue measure, and let $$Y=\prod_{i=-\infty}^{-1}X\times [0,1/2]\times\prod_{k=1}^\infty X,$$ and $A=[2/5,7/10]$. Then $\mu(A)=3/10$, but $$\pi_0^{-1}(A)\cap Y=\prod_{i=-\infty}^{-1}X\times [2/5,1/2]\times\prod_{k=1}^\infty X,$$ so $\bar\nu(\bar\pi_0^{-1}(A))=1/5$.