Let $\Omega = \{0,1\}^\infty$ and equip it with the product of discrete topologies. Let $\mathcal{F}$ be the algebra generated by the cylinder sets of $\Omega$, and let $\mathcal{B} = \sigma(\mathcal{F})$.

In a paper that I'm reading it is claimed that the following is easy to see.

If $P$ is any finitely additive probability on $(\Omega, \mathcal{B})$, then $P|_{\mathcal{F}}$, the restriction of $P$ to $\mathcal{F}$, is countably additive.

My question is:

Is there a simple proof of this claim?

The following proof seems to indicate that the claim is true, but perhaps there is a simpler argument.

Claim. If $P$ is any finitely additive probability on $(\Omega, \mathcal{F})$, then $P$ is countably additive.

Proof. Because $P$ is finitely additive, it suffices to show that $P$ is continuous above $\emptyset$. To that end let $(A_n)$ be a decreasing sequence of sets in $\mathcal{F}$ with empty intersection. We want to show that $P(A_n) \to 0$. First note that $\Omega$ is compact by Tychonoff's theorem. Also note that each $A_n$ is a closed subset of $\Omega$, because each $A_n$ is a finite union of clopen cylinders. By the finite intersection property, $\cap_{n=1}^N A_n = \emptyset$ for some $N$. Hence, $P(A_n)=0$ for $n \geq N$, and we are done.

It would be helpful if someone could address the secondary question

What, if anything, is wrong with this proof?

  • $\begingroup$ If the assertion is true then, by Caratheodory Extension Theorem, any finitely additive measure on \Omega would be countably additive. I believe this is false. $\endgroup$ – Kabo Murphy Sep 20 '17 at 6:46
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    $\begingroup$ @KaviRamaMurthy On second thought, I'm not convinced. If the proof in my question is correct, then any finitely additive $P$ on $(\Omega, \mathcal{F})$ has a unique countably additive extension to $(\Omega, \mathcal{B})$. But this does not imply that there aren't additional merely finitely additive extensions of $P$. $\endgroup$ – grndl Sep 20 '17 at 16:01

I think your proof is fine.

As another check, the same argument is used in the Kolomogorov extension theorem, which also implies the desired statement. The projection of $P$ to any $\{0,1\}^n$, call it $P_n$ (so $P_n(A) = P(A \times \{0,1\}^\omega)$, is finitely additive by assumption, and hence (trivially) countably additive since $\{0,1\}^n$ is finite. The $P_n$ obviously satisfy the Kolomogorov consistency condition and so they have a unique countably additive extension to $\{0,1\}^\omega$, call it $P'$, and by construction $P$ agrees with $P'$ on $\mathcal{F}$. (Note that, as you comment, it is still possible that $P$ and $P'$ differ off $\mathcal{F}$, so this does not prove that all finitely additive measures on $\Omega$ are countably additive, which is good because that is false.)


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