2
$\begingroup$

This is inspired by Carl Offner's reply to one of my previous questions and my previous question about marginal and joint measures.

  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!

  • 0
    @Arturo This answers actually my question. Thanks.2011-02-22

2 Answers 2

3

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.

1

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.

  • 0
    Yes, that is the reason. In particular, we usually want to have products where all factor are the same.2012-01-09