Zero Probability Is Not the Same as Impossibility
In a finite probability model with positive probability assigned to every outcome, an event of probability zero must be empty. That need not be true in general. A probability space may contain outcomes that belong to a nonempty event of probability zero. In particular, under a continuous uniform probability on an interval, each individual outcome has probability zero, even though the interval itself has probability one. Understanding this distinction is essential whenever a statement is said to hold “almost surely.”
Throughout, a probability space is a measure space \((\Omega,\mathcal{F},\mathbb{P})\) with \(\mathbb{P}(\Omega)=1\). Its elements \(\omega\in\Omega\) are outcomes, and its measurable subsets \(A\in\mathcal{F}\) are events. We use countable additivity and the consequence that if \(A\subseteq B\) are events, then \(\mathbb{P}(A)\le\mathbb{P}(B)\). No assumption that the probability space is finite or complete is made.
The empty event is always null, but a null event need not be empty. Also, an event of probability one need not equal \(\Omega\): its complement may be nonempty, provided that complement is null. The phrase “almost surely” means that the exceptional outcomes form a measurable null event; it does not mean that the property holds at every outcome.
Worked Example: A Nonempty Null Event in a Countable Space
Let \(\Omega=\{*\}\cup\mathbb{N}\), where \(\mathbb{N}=\{1,2,\ldots\}\), and let \(\mathcal{F}\) be the collection of all subsets of \(\Omega\). Define \(\mathbb{P}(\{*\})=0\) and \(\mathbb{P}(\{n\})=2^{-n}\) for \(n\ge1\). For any \(A\subseteq\Omega\), define \(\mathbb{P}(A)=\sum_{n\in A\cap\mathbb{N}}2^{-n}\).
The total mass is \(\mathbb{P}(\Omega)=\sum_{n=1}^{\infty}2^{-n}=1\), so this is a probability measure. In particular, \(\{*\}\) is a nonempty event, but \(\mathbb{P}(\{*\})=0\). The property “the outcome is a positive integer” holds on \(\mathbb{N}\), whose probability is
Thus the property holds almost surely, although it fails at the outcome \(*\). Check: \(\mathbb{N}=\Omega\setminus\{*\}\), so countable additivity gives \(\mathbb{P}(\mathbb{N})+\mathbb{P}(\{*\})=1+0=1=\mathbb{P}(\Omega)\).
Almost-Sure Properties and Exceptional Events
To express an almost-sure claim precisely, identify its exceptional event. If the claim is “the outcome belongs to a specified set \(G\),” then the property holds almost surely exactly when \(G\in\mathcal{F}\) and \(\mathbb{P}(G)=1\). Equivalently, \(G^c\) must be a measurable null event. This formulation makes measurability part of the claim: probability is defined for events in \(\mathcal{F}\), not for arbitrary subsets of \(\Omega\).
Proof. Since \(\mathcal{F}\) is a sigma-algebra, the countable union \(N=\bigcup_{k=1}^{\infty}N_k\) belongs to \(\mathcal{F}\). For every \(r\ge1\), the finite union bound gives \(\mathbb{P}(\bigcup_{k=1}^{r}N_k)\le\sum_{k=1}^{r}\mathbb{P}(N_k)=0\). The finite union is therefore null. The finite unions increase to \(N\); by continuity of probability from below, \(\mathbb{P}(N)=\lim_{r\to\infty}\mathbb{P}(\bigcup_{k=1}^{r}N_k)=0\). For the consequence, let \(N_k\) be the exceptional event for property \(k\). Outside \(\bigcup_k N_k\), none of the properties fails, and this union is null. \(\square\)
The countability condition matters. Countable additivity controls countable unions; it does not assert that an arbitrary union of null events is null. The following examples make the distinction explicit.
Worked Example: A Fixed Outcome Under Uniform Probability
Take \(\Omega=[0,1]\), let \(\mathcal{F}\) be the Lebesgue sigma-algebra on \([0,1]\), and set \(\mathbb{P}(A)=m(A)\), where \(m\) is Lebesgue measure. This is a probability space because \(m([0,1])=1\). Fix \(a\in[0,1]\). The singleton \(\{a\}\) is measurable and has measure zero: for every \(\varepsilon>0\), it is contained in an interval of length at most \(2\varepsilon\), so its Lebesgue outer measure is at most \(2\varepsilon\); letting \(\varepsilon\) decrease to zero gives \(m(\{a\})=0\).
Therefore \(\mathbb{P}(\{a\})=0\), even though the event is nonempty. Its complement is the event that the sampled point differs from \(a\), and
So the property “the outcome is not \(a\)” holds almost surely, not everywhere. Check: the two disjoint events \(\{a\}\) and \([0,1]\setminus\{a\}\) partition \([0,1]\); their probabilities \(0\) and \(1\) sum to the total probability \(1\).
Worked Example: A Countable Exceptional Set
Use the same probability space \(([0,1],\mathcal{F},\mathbb{P})\). Consider the event \(Q=\mathbb{Q}\cap[0,1]\) that the outcome is rational. It is measurable because it is countable, and it is a countable union of singleton events. Each singleton has probability zero by the preceding calculation. The proposition on countable unions therefore gives \(\mathbb{P}(Q)=0\). Hence the property “the outcome is irrational” holds almost surely.
This probability can also be checked directly. Enumerate \(Q\) without repetition as \(q_1,q_2,\ldots\). For every integer \(r\ge1\), finite subadditivity gives \(\mathbb{P}(\{q_1,\ldots,q_r\})\le\sum_{j=1}^{r}\mathbb{P}(\{q_j\})=0\). The finite sets increase to \(Q\), so continuity from below yields \(\mathbb{P}(Q)=\lim_{r\to\infty}0=0\). Thus \(\mathbb{P}([0,1]\setminus Q)=1-0=1\). Check: the rational and irrational subsets partition \([0,1]\), and their probabilities \(0\) and \(1\) add to \(1\).
Worked Example: Every Point Has Probability Zero, but the Whole Space Does Not
Still take \(\Omega=[0,1]\) with uniform Lebesgue probability. For each \(x\in[0,1]\), the event \(\{x\}\) has probability zero. Nevertheless, the union of all these events is the entire sample space:
There is no contradiction: this is an uncountable union, whereas countable additivity applies to countable families of pairwise disjoint events. In particular, one must not add the probabilities of uncountably many singletons as if countable additivity applied. The assertion “each fixed point has probability zero” is a family of individual statements; it does not say that the union of all singleton events has probability zero. Check: the union is exactly \([0,1]\), whose probability is \(m([0,1])=1\), while each individual member of the union has probability \(m(\{x\})=0\).
What “Almost Surely” Does and Does Not Allow
A useful consequence of the definition is that a property can be changed on a null event without changing whether it holds almost surely. More precisely, if two measurable sets \(A,B\in\mathcal{F}\) differ only within a null event \(N\)—that is, \(A\triangle B\subseteq N\)—then they have the same probability. Indeed, \(A\setminus B\subseteq N\) and \(B\setminus A\subseteq N\), so both differences have probability zero by monotonicity. Decomposing each set into its intersection with the other and its difference gives \(\mathbb{P}(A)=\mathbb{P}(A\cap B)=\mathbb{P}(B)\). In particular, if \(A\) has probability one, so does \(B\).
This statement concerns measurable sets and a measurable null event containing the difference. On a probability space that is not complete, a subset of a null event need not itself belong to \(\mathcal{F}\). Thus one should not silently treat every subset of a null event as an event. An almost-sure assertion is made by specifying a measurable exceptional event of probability zero.
Common mistake. From \(\mathbb{P}(\{x\})=0\) for every \(x\in[0,1]\), one might incorrectly conclude that \(\mathbb{P}([0,1])=0\) by “summing over all points.” This uses countable additivity for an uncountable family, which is not permitted. The correction is to distinguish the valid countable-union conclusion from the uncountable case: the union of the singleton events is \([0,1]\), and its probability is \(1\). Individual outcomes can be null even when the whole sample space is not.
Check Your Understanding
For each question, identify the relevant event and state whether the conclusion is pointwise or almost sure.
- In a probability space, can a nonempty event have probability zero? Give a construction or explain why not in the countable example above.
- On \([0,1]\) with uniform Lebesgue probability, what is the probability that the outcome equals a fixed \(a\in[0,1]\)? What is the probability that it differs from \(a\)?
- If \(N_k\) are null events for every positive integer \(k\), prove that the probability of \(\bigcup_{k=1}^{\infty}N_k\) is zero.
- Why does the fact that every singleton in \([0,1]\) has probability zero not imply that \([0,1]\) has probability zero?
- If each of countably many properties holds almost surely, what event must be null for all the properties to hold simultaneously?