Solution: Why Countable Additivity

Exercise: Why Countable Additivity

Part 1

For any finite set $F = \{n_1, \ldots, n_k\} \subseteq \mathbb{N}$ , finite additivity gives:

\mu(F) = \mu(\{n_1\}) + \cdots + \mu(\{n_k\}) = 0 + \cdots + 0 = 0

So $\mu$ assigns measure zero to every finite subset of $\mathbb{N}$ .

Now apply countable additivity to the partition $\mathbb{N} = \{1\} \cup \{2\} \cup \{3\} \cup \cdots$ :

\mu(\mathbb{N}) \stackrel{?}{=} \sum_{n=1}^{\infty} \mu(\{n\}) = \sum_{n=1}^{\infty} 0 = 0

But

\mu(\mathbb{N}) = \mu(\Omega) = 1

by the normalisation axiom. We have

1 = 0

— a contradiction. Therefore

\mu

cannot be countably additive while satisfying

\mu(\{n\}) = 0

for all

n

and

\mu(\mathbb{N}) = 1

However,

\mu

can be finitely additive. Any finite union of singletons has

\mu

-measure zero, and there is no contradiction with

\mu(\Omega) = 1

from finite additivity alone — finite additivity only constrains finite unions, and

\mathbb{N}

is not a finite union of singletons.

Part 2

The argument " $\mu(\mathbb{N}) = \sum_{n=1}^{\infty} \mu(\{n\}) = 0$ " is precisely the countable additivity claim applied to the partition $\mathbb{N} = \bigcup_{n \geq 1} \{n\}$ . The conclusion $\mu(\mathbb{N}) = 0$ is not a valid deduction from finite additivity — it would only follow from countable additivity, which $\mu$ does not satisfy.

This is the gap: finite additivity lets you sum finitely many disjoint events; countable additivity is the additional requirement that this also works for countably infinite disjoint collections. The two coincide on finite partitions, but diverge on infinite ones.

Part 3

Under a countably additive probability measure

\mathbb{P}

, the continuity from above property states: if

A_1 \supseteq A_2 \supseteq \cdots

and

\bigcap_n A_n = \emptyset

, then

\mathbb{P}(A_n) \to 0

Proof sketch: Write

B_n = A_n \setminus A_{n+1}

(disjoint). Then

A_1 = \bigcup_{n \geq 1} B_n

and

A_k = \bigcup_{n \geq k} B_n

. Countable additivity gives

\mathbb{P}(A_1) = \sum_{n \geq 1} \mathbb{P}(B_n) < \infty

, so the tail

\sum_{n \geq k} \mathbb{P}(B_n) \to 0

k \to \infty

, which equals

\mathbb{P}(A_k)

\square

This matters because probability limits are computed via continuity of measure. When we write $\mathbb{P}(\lim_{n \to \infty} A_n)$ and want to exchange the limit and the probability, we are using this property. Without countable additivity, limits of events have no reliable probability.

Part 4

Stochastic calculus is built on limits. Key examples:

Stopping times: $\tau = \inf\{t \geq 0 : S_t \geq K\}$ . The event $\{\tau \leq T\}$ is the union $\bigcup_{t \in [0,T]\cap\mathbb{Q}} \{S_t \geq K\}$ — a countable union (over rationals), by continuity of paths. Countable additivity ensures this event has a well-defined probability.
Convergence theorems: The dominated convergence theorem and monotone convergence theorem for integrals (expectations) rest on countable additivity of the underlying measure.
Almost-sure events: " $W_t$ has continuous paths" is not a statement about a single $t$ but about all $t$ simultaneously — a probability-one event in infinite-dimensional sample space. Its probability is 1 precisely because countable additivity extends to limits.

Finite additivity would suffice for a model with finitely many time steps and finitely many events. Continuous-time finance — where the sample space is a function space and events are limits of observable sets — requires countable additivity to give a coherent probability to any event involving a limit.

Takeaways

Finite additivity and countable additivity coincide on finite partitions but diverge on infinite ones. Countable additivity is the stronger condition and is required for probability theory to interact correctly with limits.
The continuity of measure property ( $\mathbb{P}(A_n) \to \mathbb{P}(A)$ for increasing or decreasing sequences) is a consequence of countable additivity, not of finite additivity.
In continuous-time models, nearly every interesting event is a limit — "the stock hits $K$ before $T$ ," "the portfolio never goes negative," "the integral converges." Countable additivity is not a technicality; it is the axiom that makes these events meaningful.