Measure from on product $\sigma$-algebra to on component $\sigma$-algebras

1. Given a measure $\mu$ on product $\sigma$-algebra $\prod_{i \in I} \mathbb{S}_i$ of a collection of measurable spaces $(X_i, \mathbb{S}_i), i \in I$, does there exist a measure $\mu_i$ on each component $\sigma$-algebra $\mathbb{S}_i$, s.t. their product $\prod_{i \in I} \mu_i$ is the given measure $\mu$ on the product $\sigma$-algebra?
2. If no, what are some necessary and/or sufficient conditions for the given measure $\mu$ to have such a decomposition?
3. When they exist, how to construct the component measures $\mu_i$ from $\mu$?

For example, is this a viable way by defining $$\mu_i(A_i):= \frac{\mu(A_i \times \prod_{j \in I, j\neq i} X_i)}{\prod_{j \in I, j\neq i} \mu_j(X_i)}, \forall A_i \in \mathbb{S}_i ?$$ If not, when will it become viable? ADDED: I asked this question, because obviously, the product and the division may not make sense in some cases. Also I actually made a mistake of circular definition, where I define $\mu_i$ in terms of $\mu_j, j\neq i$ which have to be defined in similar ways.

Thanks and regards!

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As was noted in your previous question math.stackexchange.com/questions/21945/…, there is an important difference between products of measurable spaces and products of measure spaces. The former always exists and is given by the categorical product, but the latter may not even exist. – Arturo Magidin Feb 22 '11 at 16:08
Your quotient may not even make sense. If one $X_i$ has infinite measure, then what does it mean to take the quotient? The product may not make sense; if one of the $X_i$ has infinite measure, and the other has measure zero, what does it meant to take their product? What is the product of an infinite number of factors? Of an uncountable number of factors? – Arturo Magidin Feb 22 '11 at 16:19
@Arturo: Thanks! As to your latest comment, I have thought about that. So Whenever possible, what are the possible ways to go from a measure on product sigma algebra to measures on component sigma algebra, s.t. their product is the given measure? – Tim Feb 22 '11 at 16:32
@Tim: You're going to have huge problems even trying. Take $X_i$ the $2$-element set, $\mathbb{S}_i$ the total $\sigma$-algebra, $\mu_i$ the counting measure, and $I$ uncountably infinite. No way to construct a product measure, because all the sets in $\prod S_i$ are uncountable, and no way to recover the counting measures from a measure on the product. Again, there may not even be a reasonable measure on the product because the projections are not measure invariant, even in the simple case of, say, $[0,1]$ with the Borel measure, and the product of two factors. – Arturo Magidin Feb 22 '11 at 16:36
@Tim: Your previous question math.stackexchange.com/questions/21945/… has an answer from Pete, which in turn has a comment directing to mathoverflow.net/questions/49426, which talks about the fact that the projections are not measure preserving (so the measure is not invariant under projections). If you pull back a measurable set, the measure of the pullback is not necessarily equal to (or every proportional to) the measure of the original set. Conversely, the measure of a projection is not necessarily well-related to the measure of the set – Arturo Magidin Feb 22 '11 at 18:26

The answer to (1) is certainly no; not every measure on a product space is a product measure, not even for a finite product. Consider for example the following: let $m$ be Lebesgue measure on $[0,1]$, let $F : [0,1] \to [0,1]^2$ be given by $F(x) = (x,x)$, and let $\mu = m \circ F^{-1}$ be the pushforward measure on the product $[0,1]^2$ (with its product $\sigma$-algebra of course). $\mu$ then is effectively 1-dimensional Lebesgue measure on the diagonal of $[0,1]$. Now $\mu$ cannot be a product measure. For suppose $\mu = \mu_1 \times \mu_2$. Let $0 < a < 1$; it's clear that we have $\mu((0,a)^2) = a > 0$ and $\mu((a,1)^2) = 1-a > 0$. Thus $\mu_i((0,a)) > 0$ and $\mu_i((a,1)) > 0$ for $i=1,2$. But $\mu((0,a) \times (a,1)) = \mu((a,1) \times (0,a)) = 0$, a contradiction.

This is best thought of in terms of probability theory: a probability measure $\mu$ on $\mathbb{R}^d$ gives the (joint) distribution of a random vector $(X_1, \dots, X_d)$. If $\pi_i : \mathbb{R}^d \to \mathbb{R}$, $i=1,\dots,d$ is the projection onto the $i$'th component, then $\mu_i = \mu \circ \pi_i^{-1}$ is the marginal distribution of $X_i$. But $\mu$ is a product measure iff $\{X_1, \dots, X_d\}$ are independent. In our example, $\mu$ is the joint distribution of $(U,U)$, where $U \sim U(0,1)$; obviously $U$ is not independent of itself.

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I answer 2 and 3 for the case of probability measures, the only measures one can sensibly take infinite products of. So all measures will be understood to be probability measures.

Let the measure $\mu_i:\mathbb{S}_i\to[0,1]$ be given by $\mu_i(S)=\mu\big(\pi_i^{-1}(S)\big)$. If $\mu$ can be written as an infinite product measure, it must be the product of the $\mu_i$, which answers 3 once we have settled the issue of when we can decompose $\mu$.

This is the case if and only if for every finite set $F\subseteq I$ and any infinite product of measurable sets $A=\prod_{i\in I}A_i$ such that $A_i=X_i$ for all $i\notin F$ (we call such sets measurable rectangles), we have $\mu(A)=\prod_{i\in F}\mu_i(A_i)$. This is clearly satisfied for product measure spaces. Conversely, measurable rectangles generate the product $\sigma$-algebra and are closed under finite intersections, so a measure on the product $\sigma$-algebra is determined by the values on these sets and the product of probability spaces always exists.

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Thanks! "probability measures, the only measures one can sensibly take infinite products of". Why is it not sensible to take infinite product of non probability measures? – Tim Jan 9 '12 at 8:45
I guess the idea is that an infinite product of positive numbers can only converge to a finite nonzero limit if they converge to $1$ (and rather quickly). Hence, the condition that $\mu_i(X_i)$ almost $1$ is necessary to get a non trivial product of the $\mu_i(X_i)$. – Did Jan 9 '12 at 8:58
Yes, that is the reason. In particular, we usually want to have products where all factor are the same. – Michael Greinecker Jan 9 '12 at 10:55