Does the Borel-Cantelli Lemma imply countable additivity?

Question

Let $(\Omega, \mathcal F, P)$ be a finitely additive probability space. If $P$ is not only finitely additive but also countably additive, then it satisfies the Borel-Cantelli Lemma:

For all sequences $A_1, A_2,...$ in $\mathcal F$, if $\sum_n P(A_n) < \infty$, then $P(\limsup_n A_n) = 0$.

I'm wondering if the converse holds as well.

Question. If $P$ (a finitely additive probability) satisfies the Borel-Cantelli Lemma, is $P$ countably additive?

Suppose that $P$ satisfies the Borel-Cantelli Lemma and that $A_1, A_2,\ldots$ is a disjoint sequence in $\mathcal F$. By finite additivity, $$\sum_n P(A_n) \leq P(\bigcup_n A_n) < \infty.$$ So, by the Borel-Cantelli Lemma $P(\limsup_n A_n)=0$, which implies $P(\liminf_n A_n^c)=1$. I tried using this fact to manipulate $$P(\bigcup_n A_n) = P(\bigcup_nA_n \cap \liminf_n A_n^c)$$ into something useful, but I wasn't able to get anywhere.

I suspect the result doesn't hold, but it seems like coming up with a counterexample (a merely finitely additive probability that satisfies the Borel-Cantelli Lemma) will be pretty difficult.

Answer 1

Here is a counterexample. Pick a nonprincipal ultrafilter $U$ on $\mathbb{N}$ and consider the finitely additive probability space $(\mathbb{N},\mathcal{P}(\mathbb{N}),P)$ where $$P(A)=\sum_{n\in A}\frac{1}{2^{n+2}}$$ if $A\not\in U$ and $$P(A)=\frac{1}{2}+\sum_{n\in A}\frac{1}{2^{n+2}}$$ if $A\in U$. (So, we have a weighted counting measure on $\mathbb{N}$ with total weight $\frac{1}{2}$, and we give an extra $\frac{1}{2}$ weight to being in $U$.) This is not countably additive since $U$ is nonprincipal so the measures of all the singletons only add up to $\frac{1}{2}$. However, I claim it satisfies the Borel-Cantelli lemma.

Indeed suppose a sequence of sets $(A_n)$ satisfies $\sum P(A_n)<\infty$. If some $k$ were in infinitely many $A_n$, then $P(A_n)$ would be at least $\frac{1}{2^{k+2}}$ for infinitely many $n$, and $\sum P(A_n)$ would diverge. Thus no $k$ is in infinitely many $A_n$, and $\limsup A_n=\emptyset$, so $P(\limsup A_n)=0$.