Exercise: Why Countable Additivity

Prerequisites: Probability Space

Problem

for all pairwise disjoint $A_1, A_2, \ldots \subseteq \mathbb{N}$.

1. Consider $\Omega = \mathbb{N} = \{1, 2, 3, \ldots\}$ with $\mathcal{F} = 2^{\mathbb{N}}$. Define $\mu(\{n\}) = 0$ for every $n \in \mathbb{N}$. Show that $\mu$ is finitely additive but not countably additive.

2. Part 1 shows that finite additivity is not enough to recover $\mu(\mathbb{N})$ from $\mu(\{n\})$. What goes wrong in the argument $\mu(\mathbb{N}) = \sum_{n=1}^{\infty} \mu(\{n\}) = \sum_{n=1}^{\infty} 0 = 0$?

3. Consider the sequence of events $A_n = \{n, n+1, n+2, \ldots\} \subseteq \mathbb{N}$ (the "tail" events). This sequence is decreasing: $A_1 \supseteq A_2 \supseteq \cdots$, and $\bigcap_{n=1}^{\infty} A_n = \emptyset$. Under a countably additive probability measure $\mathbb{P}$ on $(\mathbb{N}, 2^{\mathbb{N}})$, what must $\mathbb{P}(A_n)$ converge to as $n \to \infty$? Why does this matter for limits in probability?

4. (Conceptual) Stochastic calculus relies heavily on taking limits of events (e.g., "$S_t > K$ for some $t \in [0, T]$"). Give an informal argument for why countable additivity — not just finite additivity — is the right axiom for a theory that involves limits.

Hint

For part 1, compute $\mu$ of a finite set using finite additivity, then try to apply the claimed formula to $\mathbb{N}$ itself.