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.