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Given two measure spaces $(\Omega_i, \mathcal{F}_i, \mu_i), i=1,2$, does there always exists a measure preserving mapping $(\Omega_1, \mathcal{F}_1, \mu_1) \to (\Omega_2, \mathcal{F}_2, \mu_2)$?

One necessary condition is that $\mu_1(\Omega_1) = \mu_2(\Omega_2)$. But is that also sufficient?

I am in the middle of understanding what the Kolmogorov extension theorem is all about.

  • Some version states without a stochastic process, i.e. for any consistent finite dimensional distributions, under some condition, there exists a measure on the space of the sample paths s.t. the finite dimensional distributions are its marginal distributions.
  • Some version states in terms of a stochastic process, i.e. for any consistent finite dimensional distributions, under some condition, there exists a stochastic process whose measure on the space of the sample paths has the finite dimensional distributions as its marginal distributions.

Obviously the latter version implies the former version, but I am not sure the former can imply the latter. That serves as the source of my questions.

BTW, are the measures in KET all probability measures, not general ones? I saw at the end of this note, the measures in its version of KET are required to be Baire measures, which I think may not be probability measures?

Thanks and regards!

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"These are not droids you are looking for"

Both statements you cited are equivalent and are not very relevant to the existence of measure-preserving maps between two fixed probability spaces. Why?

Suppose, you are looking for a stochastic process $X$ taking values in the space $E$ with certain pre-defined f.d.d. Then you can consider $X$ to be a random element $X:\Omega_1\to\Omega_2$ where $\Omega_2 = E^{[0,\infty)}$ is the space of trajectories of the process which is fixed. What KET states, is that given a collection of f.d.d. satisfying two assumptions there exists a unique probability measure $\mathsf P$ on $\Omega_2$ which marginals coincide with f.d.d. This does not tell you about the existence of $X$? It actually does: initially, $\Omega_1$ is just "some" sample space and is not fixed. So what we can do is to take $\Omega_1 = \Omega_2$, as for the latter we don't have any choice (it's a space of trajectories), and then we set $X =\mathrm{id}_{\Omega_2}$.

Take-away point: stochastic process when considered as a single process is just a distribution on the space of trajectories - the so-called "canonical probability space". The correspondent sequence of random variables can be easily taken to be an identity map. More generally, given any probability space $(\Omega,\mathscr F,\mathsf P)$ there always exists a random variable $\xi$ with a distribution $\mathsf P$, e.g. $\mathrm{id}_\Omega$. More detailed discussion is here - see also the MO link there.

However, when you are interested in considering at least two stochastic processes at the same time it's always necessary to specify the common probability space for all processes in focus, which though may differ from the canonical probability space for each of them.

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  • $\begingroup$ Thanks for pointing out the thing I misunderstood about KET and the interesting links! I didn't realize the underlying space is not specified in KET version using a stochastic process. So my original question of given fixed domain and codomain asking for a measure preserving mapping is not a concern in KET. But as stand-alone, what would you think about the original question? $\endgroup$ – Tim Jan 16 '13 at 18:02
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    $\begingroup$ @Tim take the first space to be a singleton and the seance one to be a unit interval with the Lebesgue measure $\endgroup$ – Ilya Jan 16 '13 at 21:03
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Surely the answer is no. For example, take $\Omega_i$ to be the integers with all subsets measurable, and define $\mu_1(\{x\}) = 2$, $\mu_2(\{x\}) = 1$ for any $x$. Then $\mu_2(\{0\}) = 1$, but $\mu_1(f^{-1}(\{0\}))$ is even.

Also, $\mu_i(\Omega_i)$ is infinite for each $i$, so the total measures match.

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  • $\begingroup$ Thanks! (1) How is a point measure defined? Can "a point would have to be mapped to a point that gives those singleton sets equal measure" be rephrased more precisely? (2) Does the version of KET without using a stochastic process need weaker condition than the version using a stochastic process? $\endgroup$ – Tim Jan 16 '13 at 17:03
  • $\begingroup$ Sorry, I know nothing about KET, just some very basic measure theory. See if this helps: $\Omega_1$ is the integers, $F_1$ is the set of all subsets of the integers, and $\mu_1$ takes every set to the size of that set. Check that this defines a measure. Similarly for $i = 2$, except that $\mu_2$ is twice as large. Then pick a point (say $0$) and think about the measure of that point and the measure of its image. $\endgroup$ – Hew Wolff Jan 16 '13 at 17:10
  • $\begingroup$ Thanks! A measure preserving mapping $f$ from $(\Omega_1, \mathcal{F}_1, \mu_1) \to (\Omega_i, \mathcal{F}_i, \mu_i)$ is a measurable mapping s.t. $\forall A \in \mathcal{F}_2$, $\mu_1(f^{-1}(A)) = \mu_2(A)$. So I think that the empty domain example doesn't work, because $\mu_1(f^{-1}(\Omega_2)) = \mu_1(\emptyset ) = 0 \neq \mu_2(\Omega_2)$ $\endgroup$ – Tim Jan 16 '13 at 17:53
  • $\begingroup$ Right you are, I'll remove that. $\endgroup$ – Hew Wolff Jan 16 '13 at 20:36
  • $\begingroup$ The example in the first paragraph is based on $\forall A \in \mathcal{F}_1, \mu_2(f(A)) = \mu_1(A)$, which might not work either for the definition of a measure preserving mapping, I guess? $\endgroup$ – Tim Jan 16 '13 at 20:52

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