Given an increasing sequence of finite sigma algebras $\mathcal{F}_n$ on the measurable space $(\Omega,\mathcal{F})$ and a consistent sequence of probabilities $P_n$ such that $P_{n+1}$ restricted to $\mathcal{F}_n$ is the same as $P_n$, there exists a probability $P$ on $\sigma(\bigcup_{k=1}^{\infty} \mathcal{F}_n)$ such that $P\big|_{\mathcal{F}_n}=P_n$

Why is $P$ well defined on $\cup_{n=1}^{\infty} \mathcal{F_k}$ I can easily show finite additivity of $P$ on $\cup_{n=1}^{\infty} \mathcal{F}_k$ using consistency above , but how can I show countable additivity?(In general I cannot use a large enough $m \in \mathbb{N}$ such that I use the countable additivity of the corresponding $P_m$ which is given to be a probability)

The author of the book I am reading (Ikeda and Watanabe) further claims that we can show that $P$ is indeed a probability measure on $\sigma(\cup_{k=1}^{\infty} \mathcal{F_k})$ using the Hopf's extension theorem. Can somebody give me a hint or some details?

  • $\begingroup$ An element $A\int \bigcup_k F_k$ must belong to some $F_n$. Therefore $P_n(A)=P_{n+1}(A)=...$. This shows that $P$ is well defined on $A$. $\endgroup$ – YAlexandrov Mar 12 '18 at 9:42
  • $\begingroup$ @YAlexandrov i dont understand your notation $\endgroup$ – user3503589 Mar 12 '18 at 9:48
  • $\begingroup$ Ah, I wrote \int instead of \in $\endgroup$ – YAlexandrov Mar 12 '18 at 9:49
  • $\begingroup$ An element $A\in \bigcup_{k=1}^{\infty}F_k$ must belong to some $F_n$. Therefore $P_n(A)=P_{n+1}(A)=...$. This shows that $P$ is well defined. This is how it should have looked. $\endgroup$ – YAlexandrov Mar 12 '18 at 9:50
  • 1
    $\begingroup$ en.wikipedia.org/wiki/Kolmogorov_extension_theorem $\endgroup$ – Did Mar 12 '18 at 9:52

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