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I would like to prove the following, using the standard Kolmogorov extension theorem (e.g. http://en.wikipedia.org/wiki/Kolmogorov_extension_theorem):

Let $(\Omega, \mathscr{F}, P)$ denote our underlying probability space. Let $(T, \mathscr{G})$ be a nice topological measure space (i.e. $\mathscr{G}$ arises from the topology defined on T). Suppose that $(T_n)_{n=1}^\infty$ is a nested sequence of subsets of $T$, increasing to $T$. Consider a sequence $(X_n)_{n=1}^\infty$ of random variables, where each $X_n$ takes values in $T_n$.

Suppose moreover that the sequence $(X_n)_{n=1}^\infty$ is 'consistent' in the following sense:

$$ \forall n \in \mathbb{N} \quad \forall A \in \mathscr{G} \quad P(X_{n+1} \in A | X_{n+1} \in T_n) = P(X_n \in A). $$

Then there exists a measure $\mu$ on $(T, \mathscr{G})$ such that

$$ \forall n \in \mathbb{N} \quad \forall A \in \mathscr{G} \quad \mu(A \cap T_n) = P(X_{n} \in A).$$

My questions are: 1. Is this true; if so, how to prove it? 2. I think it may be necessary to assume that $T$ is sigma-finite and that the $T_n$'s have finite measure. Is this so? 3. What topological properties should the space $T$ have?

Many thanks for your help.

Frank.

$\textbf{EDIT}$ I have now proved the required result, taking $(T, \mathscr{G})$ to be any measurable space (i.e. not necessarily topological as above). I do not assume this space is sigma-finite either.

$\textbf{EDIT}$ I have spotted and error in my proof, and now think that actually the result may be false (though all I can actually conclude is that my basic attempt at a proof is wrong!). This is how my proof began:

$\textbf{PROOF}$ Define $\mu(A) := \sum_{n \in \mathbb{N}} {P}(X_n \in A \cap (T_n \setminus T_{n-1})$. It is easy to show that $\mu$ is a measure on $(T, \mathscr{G})$.

It remains to show the desired property. By definition $ \forall n \in \mathbb{N} \quad \forall A \in \mathscr{G} \quad$ $$ \mu(A \cap T_n) = \sum_{k \in \mathbb{N}} {P}(X_k \in A \cap T_n \cap (T_k \setminus T_{k-1})) = \sum_{k \leq n} {P}(X_k \in A \cap (T_k \setminus T_{k-1})) .$$

But
\begin{align*}\sum_{k \leq n} {P}(X_k \in A \cap (T_k \setminus T_{k-1}))& = \sum_{k \leq n} {P}(X_n \in A \cap (T_k \setminus T_{k-1})| X_n \in T_k)\\ &= \sum_{k \leq n} \frac{{P}(X_n \in A \cap (T_k \setminus T_{k-1}))}{{P}( X_n \in T_k)}\\ &\geq \sum_{k \leq n} {P}(X_n \in A \cap (T_k \setminus T_{k-1}))\\ &= {P}(X_n \in A) . \end{align*} where the first equality follows by induction, give our assumption on the conditional probability distributions of the $X_n$.

However, it is not possible to prove the reverse inequality as required.

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  • $\begingroup$ The proof of the extension theorem in generality requires a kind of inner regularity of measures. Classical results in measure theory give that on nice enough topological spaces, all Borel measures will have have sufficient regularity. If you want sharp conditions, you're in for a bit of work. If you want sufficient conditions, you can take your topological spaces to be Polish or Hausdorff. If Hausdorff, you need to add the assumption of tightness to the measures you define, since not all probability measures are necessarily tight in Hausdorff spaces. $\endgroup$ Mar 15, 2014 at 14:02

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