Measure from on product $\sigma$-algebra to on component $\sigma$-algebras This is inspired by Carl Offner's reply to one of my previous questions and my previous question about marginal and joint measures.


*

*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?

*If no, what are some necessary and/or
sufficient conditions for the given
measure $\mu$ to have such a
decomposition?

*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!
 A: 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.
A: 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.
