# Does the Borel-Cantelli Lemma imply countable additivity?

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.

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 $$P(\limsup A_n)=0$$.
• Actually, this example makes it seem like it's fairly easy for a merely finitely additive probability (on $\mathbb N$) to satisfy BCL. All we need is for singletons to get positive measure. I'll have to think about how this carries over to uncountable spaces.
• Right, the Borel-Cantelli lemma is trivial, not interesting, for discrete probability spaces. What makes it interesting and useful is that it works even when the $A_n$ aren't built out of atoms. Jun 18 '20 at 4:11