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  1. Given a measure space $(X, \mathcal F, \mu)$, it is always possible to construct a measurable mapping defined from some other measure space and with $X$ as its codomain, such that the measure induced by the mapping is the measure $\mu$ on the codomain.

    Suppose we further require the measurable mapping to satisfy that given some subset $A$ of $X$ ($A$ is not necessarily measurable, i.e. may not be in $\mathcal F$), the mapping maps into $A$ a.e.. Then the existence of such a mapping requires that there exists a $B \in \mathcal F$ such that $B \subseteq A$ and $\mu(X - B) = 0$ (correct me if I am wrong). I was wondering if that necessary condition is also sufficient for existence of such a mapping?

    Is it correct that existence of such a $B$ is completely determined by the measure space $(X, \mathcal F, \mu)$? Among all the measurable mappings that can induce $\mu$, either all of them map into $A$ a.e., or neither of them will map into $A$ a.e.? It is impossible that some of them map into $A$ a.e. and some don't?

  2. As an example and also source of my above question, let's look at construction of a Brownian motion:

    Given finite dimensional distributions of a Brownian motion, there exists a unique probability measure $P$ on the sample path space $\mathbb{R}^T$ where $T:=[0, \infty)$ with its Borel sigma algebra, by Kolmogorov extension theorem.

    A Brownian motion also requires its sample path to be continuous a.e.. Is existence of a Brownian motion equivalent to that the set $A$ of continuous functions from $T$ to $\mathbb{R}$ contains a subset which has probability $1$ under $P$? (Note $A$ is not measurable wrt the Borel sigma algebra over $\mathbb{R}^T$.)

    Is it possible that there exists a stochastic process with $P$ being its law, but not continuous a.e.?

Thanks and regards!

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1 Answer 1

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  1. This is a sufficient condition, but it is not necessary. A necessary and sufficient condition for the existence of such a mapping is that $\mu(S)=0$ for all $S \in\cal F$ with $S\cap A=\emptyset$. Also, given a mapping that induces $\mu$, and given that this condition holds, it is not predetermined whether or not the mapping is a.s. into $A$. It may be that some mappings are a.s. into $A$ and some are not.

  2. You can construct processes which have the same finite-dimensional distribution as Brownian motion but which are a.s. continuous nowhere. For example, let $(B_t)_{t\ge 0}$ be the usual Brownian motion, which is a.s. continuous everywhere, and let $U$ be uniformly distributed on $[0,1]$. Then pick a dense set $M$ of measure $0$ and a badly behaved function $f$ and set $$B'_t:=\left\{\begin{array}{ll} B_t+f(t),&\qquad t+U\in M,\\ B_t,&\qquad t+U\notin M.\end{array}\right.$$ For any choice $(t_1,\dots,t_k)$ of a finite number of times, $U$ will a.s. not be in any of $M-t_1$, $\dots$, $M-t_k$, so the finite-dimensional distributions of $B_t$ and $B'_t$ are identical. However, if $f$ has been chosen so that $f \chi_M$ is discontinuous everywhere on ${\Bbb R}$, $B'_t$ will a.s. be discontinuous everywhere on ${\Bbb R}_{\ge 0}$.

    The moral is that you cannot construct Brownian motion directly by using the Kolmogorov extension theorem on the product space over all real times. It's possible to construct it by using the Kolmogorov extension theorem on the product space over a countable dense set of times and then extending the motion by continuity.

Re your additional questions:

(1), (4): Your condition is stronger. It implies that all mappings inducing $\mu$ are a.s. into $A$. My condition says that a $\mu$-inducing mapping a.s. into $A$ exists, not that all $\mu$-inducing mappings are a.s. into $A$.

(2) Correct. A.s. sample path continuity is not determined by finite-dimensional distributions, or, equivalently, by the measure on the product $\sigma$-algebra.

(3) Applying the Kolmogorov extension theorem to ${\Bbb R}^{[0,\infty)}$ gives an underdetermined random variable, since, in the product space over all nonnegative real times, there are uncountably many times, but any event in the product $\sigma$-algebra can depend on only countably many times.

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Thanks! (1) For the first part, your condition is sufficient and necesary. I think your condition is completely determined by the measure space. Then why "It may be that some mappings are a.s. into A and some are not"? (2) So being continuous a.s. cannot be determined by the law completely? (3) What makes it fail to construct on the product space over all real times, but succeed on the product space over a countable subset? –  Tim Feb 2 '13 at 21:29
    
(4) In the first part, why is your condition necessary and sufficient, while mine is just sufficient? –  Tim Feb 2 '13 at 22:03
    
@Tim - I replied to these questions above. –  David Moews Feb 5 '13 at 0:09
    
THanks! I now understand your points. For (3), how is "an underdetermined random variable" defined? Why "any event in the product σ-algebra can depend on only countably many times"? –  Tim Feb 5 '13 at 22:27
    
For (3), I was using the word "undeterdetermined" in an informal sense, so it has no exact definition. The second statement is true because the events depending on only countably many times contain all the generating events and are closed under the $\sigma$-algebra operations. –  David Moews Feb 5 '13 at 22:42

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