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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.

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  • $\begingroup$ Can you solve this when there are just two factors? For some $Y$ there exists (one or more) such measures; for some $Y$ there is no such measure. $\endgroup$ – GEdgar Mar 25 '18 at 1:37
  • $\begingroup$ @GEdgar Please see my edit $\endgroup$ – K.Power Mar 25 '18 at 3:00
  • $\begingroup$ If $Y$ is the "diagonal" $\Delta$ then of course measure $\nu$ exists. The set $\Delta$ consists of all the "constant" functions $\mathbb Z \to X$. As I said before: Why not solve this in the case of two factors first? Every projection having full measure is necessary, but not sufficient, for $\nu$ to exist. Even when there are only two factors. $\endgroup$ – GEdgar Mar 25 '18 at 12:14
  • $\begingroup$ @GEdgar my idea is to use the Caratheodory extension theorem. We can define $\nu$ on the semi algebra consisting of all finite intersections of inverse images of $X$ under the projections in the natural way as a product of $\mu$. If $Y$ is such that $\nu$ is countably additive on the semi algebra then I believe we will be done. I have not yet figured out what are the sufficient conditions for this to be true though. Is this the correct approach at least? $\endgroup$ – K.Power Mar 25 '18 at 15:17
  • $\begingroup$ A nice example in 3 dimensions. In the cube $[0,1]^3$, with Lebesgue measure on each factor. Use the sphere inscribed in that cube with ordinary surface-area measure. Then the projections onto the one-dimensional axes are all uniform. $\endgroup$ – GEdgar Mar 25 '18 at 16:44

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