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.