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Let $x_n$ be a sequence of real numbers in $[0,1]$ such that $\sum_n x_n < 1$.

Does there exist a finitely additive probability measure $P$ on $(\mathbb{N}, 2^\mathbb{N})$ such that $P(\{n\}) = x_n$ for all $n \in \mathbb{N}$?

It seems very likely to me that the answer is yes because it is well-known that there exists a finitely additive $\mu$ on $(\mathbb{N}, 2^\mathbb{N})$ that vanishes on singletons. $\mu$ embodies the most radical failure of countably additivity that is possible with $\sum_n \mu(\{n\}) = 0 \neq 1 = \mu(\mathbb{N})$. So surely less radical failures of countable additivity are possible too. That is my intuition anyway, but I'm not sure how to prove it.

The only idea I have is the following. Define $$P_n = \sum_{j=1}^n x_n \delta_n + \Big(1-\sum_{j=1}^nx_n\Big)\mu.$$ Since $P_n$ is a convex combination of finitely additive probabilities, it too is a finitely additive probability. Moreover, $P_n$ converges pointwise to a set function $P$ such that $P(\{n\})=x_n$. Since finitely additive probabilities are closed under pointwise convergence, $P$ is the desired finitely additive probability.

Is that an okay proof? (It occurs to me that it might be better to describe the convergence as weak*. I hope someone can comment on this.)

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Your proof is fine. There's an even easier proof, though: just directly define $$P=\sum_{n=1}^\infty x_n \delta_n + \Big(1-\sum_{n=1}^\infty x_n\Big)\mu.$$ Note that the first term is just an ordinary (countably additive measure), a weighted counting measure. So this $P$ is just the sum of two finitely additive measures, and is thus a finitely additive measure.

In fact, every example must be of this form. If $P$ is a finitely additive probability measure with $P(\{n\})=x_n$, then $$\mu=\frac{P-\sum_{n=1}^\infty x_n \delta_n}{1-\sum_{n=1}^\infty x_n}$$ is a finitely additive probability measure which vanishes on singletons. (Note that $\mu$ is nonnegative since $P(A)\geq P(A\cap\{1,\dots,n\})$ for each $n$ and taking the limit as $n\to\infty$ gives $P(A)\geq \sum_{n=1}^\infty x_n\delta_n(A)$.)

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