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 $P(\limsup A_n)=0$.

  • $\begingroup$ Very nice, thanks! $\endgroup$
    – aduh
    Jun 18 '20 at 3:44
  • $\begingroup$ 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. $\endgroup$
    – aduh
    Jun 18 '20 at 4:06
  • $\begingroup$ 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. $\endgroup$ Jun 18 '20 at 4:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.