Let $(X,\Sigma, \mu)$ be a standard Borel space probability space, i.e $\mu(X)=1$. Consider the space $(X^\mathbb Z, \Sigma_p)$, where $\Sigma_p$ is the standard $\sigma$ algebra on the product space generated by the coordinate projections $\pi_n:X^{\mathbb Z}\to X$. Now let $Y\subset \Sigma_p$, and give it the standard subspace $\sigma$-algebra $\mathcal S=\Sigma_p\cap Y$. It is fairly easy to see that the restriction of each coordinate map $\pi_n|_Y$ is measurable w.r.t $\mathcal S$. What I am interested in though, is
Does there exist a probability measure $\nu$ on $\mathcal S$ such that each restricted projection is $\pi_n|_Y$ is $\mu/\nu$ measure preserving?
If $Y$ is itself just a product of subsets of $X$, then the answer is yes due to the Kolmogorov Extension theorem (in fact there are other applicable extension theorems too I believe, so pick your favourite). What is not so clear to me is what happens when $Y$ is not itself just a product space. My first consideration was the measure $\nu(A)=m(A)/m(Y)$, where $m$ is the measure on $\Sigma_p$ we get via the Kolmogorov extension theorem. Unfortunately the restricted projections do not need to be measure preserving with respect to this measure, as I figured out after asking here.
Important Edit: I have been made aware of a trivial counterexample. Consider $X=[0,1]$ with Lebesgue measure $\lambda$, and let us focus only on the simple case of $X\times X$. Let $Y=X\times\{0\}$. This is clearly measurable, but $\pi_2|_Y=0$, which implies that our desired $\nu$ on $Y$ cannot exist, because $\pi_2|_Y^{-1}\{0\}=Y$.
I had hoped that maybe demanding that $\mu(\pi_n(Y))>0$ would be a sufficient condition ensuring that such a measure existed, but on further reflection I realised if we let $X$ be as above and consider $Y=X\times [0,1/2]\subset X\times X$ we run into a problem, because $\pi_2|_Y^{-1}(1/2,1]=\emptyset$. This has made me realize that my above claim about the Kolmogorov extension was entirely incorrect, because obviously the consistency conditions are not satisfied if we take any old product of subsets.
I have posted a bounty though, and I am still interested in the subtleties of this problem, so let me ask a more subtle question:
If $Y$ is a measurable proper subset of $X^\mathbb{Z}$ under what conditions does there exist a measure $\nu$ on $\mathcal S$ making the restriction of each coordinate projection measure preserving?
I believe that a necessary condition is requiring that image of each projection has full measure, but I'm not sure whether this is sufficient.