Probability as a Continuous Set Function

May 2026 · 14 min read

Continuity is the link between “finite” probability and “limit” probability. Concretely: if a sequence of events grows toward some limiting event $E$, do their probabilities tend to $P(E)$? The answer is yes, and that single fact — called continuity of probability — is hiding inside countable additivity. In fact, the two are equivalent.

Motivation

Toss a coin repeatedly. Let $E_n$ = "at least one head in the first $n$ tosses". Then $E_1 \subseteq E_2 \subseteq E_3 \subseteq \cdots$ — if you've seen a head by toss $n$, you've also seen one by toss $n + 1$. The natural “limit” event is $E$ = “a head ever appears” = $\bigcup_n E_n$.

We have $P(E_n) = 1 - (1/2)^n$. Intuitively $P(E)$ should be $\lim_n P(E_n) = 1$. The continuity property is the rigorous statement of this intuition.

Increasing and Decreasing Sequences of Events

For increasing sequences, $\lim E_n$ is what eventually accumulates: an outcome belongs to the limit iff it is in some $E_n$. For decreasing sequences, $\lim E_n$ is what survives: an outcome belongs iff it is in every $E_n$.

The Continuity Theorem

Proof (Increasing Case)

Given $E_1 \subseteq E_2 \subseteq \cdots$. Build a disjoint sequence with the same union:

• $F_1 = E_1$,
• $F_n = E_n \setminus E_{n-1} = E_n \cap E_{n-1}^c$ for $n \ge 2$.

The $F_n$ are pairwise disjoint (if $m < n$, $F_m \subseteq E_m \subseteq E_{n-1}$ but $F_n \cap E_{n-1} = \emptyset$), and

$\displaystyle \bigcup_{i=1}^{n} F_i = E_n, \qquad \bigcup_{i=1}^{\infty} F_i = \bigcup_{i=1}^{\infty} E_i = \lim_n E_n.$

Now apply countable additivity (Axiom 3):

$\displaystyle P\!\left(\lim_n E_n\right) = P\!\left(\bigcup_{i=1}^{\infty} F_i\right) = \sum_{i=1}^{\infty} P(F_i) = \lim_{n \to \infty} \sum_{i=1}^{n} P(F_i) = \lim_{n \to \infty} P(E_n).$

The third equality is the definition of an infinite series; the fourth uses finite additivity to recognise the partial sum as $P(E_n)$. Done.

Proof (Decreasing Case)

If $E_1 \supseteq E_2 \supseteq \cdots$, the complements $E_n^c$ form an increasing sequence with limit $\bigcup_n E_n^c = (\bigcap_n E_n)^c$ (De Morgan). Apply the increasing case to $\{E_n^c\}$ and use $P(E_n) = 1 - P(E_n^c)$ to conclude.

Continuity ⇔ Countable Additivity

The proof above showed that countable additivity implies continuity. The converse is also true: if a finitely additive set function is continuous (on monotone sequences) then it is countably additive.

To see why, write any disjoint sequence $G_1, G_2, \ldots$ as the limit of partial unions $H_n = G_1 \cup G_2 \cup \cdots \cup G_n$ (which is increasing). Continuity gives $P(\bigcup_n G_n) = \lim_n P(H_n)$, and finite additivity gives $P(H_n) = \sum_{i=1}^n P(G_i)$. So

$\displaystyle P\!\left(\bigcup_{n} G_n\right) = \lim_{n} \sum_{i=1}^{n} P(G_i) = \sum_{i=1}^{\infty} P(G_i),$

which is exactly Axiom 3. Hence continuity is the “dynamic” face of countable additivity.

Visual: Probability of Nested Sets

Step through an increasing and then a decreasing sequence of events and watch the probability tend to its limit.

Examples That Use Continuity

Example 1. The First Head Eventually Comes Up

Let $E_n$ = "head appears in first $n$ tosses". Increasing, with $P(E_n) = 1 - 2^{-n}$. By continuity,

$P(\text{head ever}) = \lim_n (1 - 2^{-n}) = 1.$

So the event “a head occurs at some point” has probability $1$ in fair coin tossing, even though no individual toss is guaranteed to be a head.

Example 2. The Coin Never Comes Up Heads

Let $F_n$ = "first $n$ tosses are all tails". Decreasing, with $P(F_n) = 2^{-n}$. Limit event: “every toss is a tail” = $\bigcap_n F_n$. By continuity,

$P(\text{never any head}) = \lim_n 2^{-n} = 0.$

Example 3. The St Petersburg Setup

Toss a fair coin until you see the first head. Let $T$ be the toss number on which it appears. Then $T$ is finite with probability $1$:

$P(T < \infty) = P\!\left(\bigcup_n \{T \le n\}\right) = \lim_n P(T \le n) = \lim_n (1 - 2^{-n}) = 1.$

Without continuity we could not even justify the everyday claim “eventually we see a head”.

General Sequences (limsup, liminf)

For an arbitrary sequence $\{E_n\}$ — not necessarily monotone — we still get useful limiting events:

• $\displaystyle \limsup_n E_n = \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} E_k$, the event that infinitely many $E_n$ occur.
• $\displaystyle \liminf_n E_n = \bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty} E_k$, the event that all but finitely many $E_n$ occur.

The Borel–Cantelli lemmas (covered later in Ross) bound $P(\limsup_n E_n)$ in terms of $\sum_n P(E_n)$. Their proofs lean on the continuity property we just established.

Common Mistakes

• Applying continuity to a non-monotone sequence directly. You need either $E_n \uparrow$ or $E_n \downarrow$. For general sequences, decompose into liminf/limsup.
• Confusing $\bigcup_n E_n$ with $\sum_n P(E_n)$. The former is a set; the latter is a number that matches it only when the $E_n$ are disjoint.
• Assuming a probability-$0$ event is the same as the impossible event. “The coin never comes up heads” has probability $0$ but is a perfectly real outcome of the experiment.
• Using continuity for finitely additive but not countably additive measures — pathological, doesn't work, and not relevant for this course.

Practice

1. Let $E_n = (0, 1 - 1/n]$ inside $S = [0, 1]$ with the uniform probability $P(I) = $ length($I$). Check increasing, find $\lim_n E_n$ and $\lim_n P(E_n)$. Verify continuity.
2. Let $F_n = (0, 1/n]$. Check decreasing, find $\lim_n F_n$ and $\lim_n P(F_n)$. Verify continuity.
3. Toss a fair coin infinitely often. Show that the probability of seeing infinitely many heads is $1$. (Hint: complement plus continuity.)
4. Prove the decreasing case of the continuity theorem directly, without invoking the increasing case via complements.
5. Construct a finitely additive but not countably additive probability on $\mathbb{N}$ — the “density” or “Banach” probability. Show it fails continuity. (Hard; advanced reading.)
6. Prove the first Borel–Cantelli lemma: if $\sum_n P(E_n) < \infty$, then $P(\limsup_n E_n) = 0$.