Limits of Sequences of Events
For an event sequence, the natural set limits are its union and intersection when the events are nested. If \(E_1\subseteq E_2\subseteq\cdots\), an outcome belongs to at least one \(E_n\) exactly when it belongs to \(\bigcup_{n=1}^{\infty}E_n\). If \(A_1\supseteq A_2\supseteq\cdots\), an outcome belongs to every \(A_n\) exactly when it belongs to \(\bigcap_{n=1}^{\infty}A_n\). Continuity of probability says that the probabilities of these limits are the corresponding limits of the probabilities.
Throughout, \((\Omega,\mathcal{F},\mathbb{P})\) is a probability space: \(\mathcal{F}\) is a sigma-algebra of events, \(\mathbb{P}\) is a countably additive measure on \(\mathcal{F}\), and \(\mathbb{P}(\Omega)=1\). Thus \(\mathbb{P}\) is finite, since \(0\le\mathbb{P}(E)\le1\) for every event \(E\in\mathcal{F}\). The space need not be complete. All event equalities below are exact set equalities, and all probability equalities are exact, not merely almost sure statements.
Countable additivity also gives finite additivity. Indeed, for disjoint events \(C_1,\dots,C_k\), apply countable additivity to \(C_1,\dots,C_k,\varnothing,\varnothing,\dots\). In particular, if \(C\) is an event, then \(\Omega=C\cup C^c\) is a disjoint union and \(\mathbb{P}(C^c)=1-\mathbb{P}(C)\). These facts will be used in both continuity results.
Continuity from Below
Proof. Define \(D_1=E_1\), and for \(n\ge2\) define \(D_n=E_n\setminus E_{n-1}\). The sets \(D_n\) are events and are pairwise disjoint: if \(j<n\), then \(D_j\subseteq E_j\subseteq E_{n-1}\), whereas \(D_n\) is disjoint from \(E_{n-1}\). For every \(N\ge1\), the disjoint union \(D_1\cup\cdots\cup D_N\) equals \(E_N\). This follows by adding, at each step, precisely the part of \(E_n\) not already in \(E_{n-1}\). Also, \(\bigcup_{n=1}^{\infty}D_n=\bigcup_{n=1}^{\infty}E_n\): each \(D_n\) is contained in \(E_n\), and each point in the union of the \(E_n\) belongs to its first such event and hence to one of the \(D_n\). Countable additivity now gives $$ \mathbb{P}\left(\bigcup_{n=1}^{\infty}E_n\right) =\sum_{n=1}^{\infty}\mathbb{P}(D_n) =\lim_{N\to\infty}\sum_{n=1}^{N}\mathbb{P}(D_n) =\lim_{N\to\infty}\mathbb{P}(E_N). $$ The series has nonnegative terms, so its partial sums converge to its sum, with the sum finite here because it is at most \(1\). If \(E_1=\varnothing\), the argument still applies with \(D_1=\varnothing\). \(\square\)
The proof identifies why nesting matters: it turns the newly added portions of the events into disjoint pieces. Their probabilities can then be added by countable additivity.
Worked Example: An Increasing Sequence with a Geometric Remainder
Suppose \((E_n)\) is increasing and \(\mathbb{P}(E_n)=1-2^{-n}\) for every \(n\ge1\). Find the probability of \(\bigcup_{n=1}^{\infty}E_n\), and check the calculation from the disjoint increments.
Since the events are increasing, continuity from below applies. Substituting the stated probabilities gives $$ \mathbb{P}\left(\bigcup_{n=1}^{\infty}E_n\right) =\lim_{n\to\infty}(1-2^{-n})=1. $$ For a check, the first increment has probability \(\mathbb{P}(E_1)=1-2^{-1}=1/2\). For \(n\ge2\), the increment \(D_n=E_n\setminus E_{n-1}\) has probability $$ \mathbb{P}(D_n)=\mathbb{P}(E_n)-\mathbb{P}(E_{n-1}) =(1-2^{-n})-(1-2^{-(n-1)})=2^{-n}. $$ Thus the probability of the union, computed by adding disjoint increments, is $$ \frac12+\sum_{n=2}^{\infty}2^{-n} =\frac12+\frac{2^{-2}}{1-2^{-1}} =\frac12+\frac12=1. $$ At stage \(N\), the same calculation gives \(\frac12+\sum_{n=2}^{N}2^{-n}=1-2^{-N}\), which agrees with the specified value of \(\mathbb{P}(E_N)\). Both computations give union probability \(1\).
Continuity from Above
Proof. The complements \(A_n^c\) form an increasing sequence, because \(A_{n+1}\subseteq A_n\) implies \(A_n^c\subseteq A_{n+1}^c\). De Morgan's law gives $$ \bigcup_{n=1}^{\infty}A_n^c=\left(\bigcap_{n=1}^{\infty}A_n\right)^c. $$ By continuity from below and the complement formula, $$ 1-\mathbb{P}\left(\bigcap_{n=1}^{\infty}A_n\right) =\mathbb{P}\left(\bigcup_{n=1}^{\infty}A_n^c\right) =\lim_{n\to\infty}\mathbb{P}(A_n^c) =\lim_{n\to\infty}\bigl(1-\mathbb{P}(A_n)\bigr). $$ The sequence \(\mathbb{P}(A_n)\) is nonincreasing and bounded below by \(0\), so its limit exists. Subtracting the displayed equality from \(1\) proves the result. This argument includes the cases where the intersection is empty or equals \(\Omega\). \(\square\)
Equivalently, the probabilities of a decreasing sequence of events converge because they are nonincreasing and bounded below, and continuity from above identifies their limit as the probability of the intersection. The probability-space assumption supplies the finite total measure needed for the complement argument.
Worked Example: A Decreasing Sequence with a Nonzero Intersection
Suppose \((A_n)\) is decreasing and \(\mathbb{P}(A_n)=\frac12+\frac{1}{2n}\) for every \(n\ge1\). Find the probability of \(\bigcap_{n=1}^{\infty}A_n\), and verify it by accounting for the portions removed at each step.
Continuity from above gives $$ \mathbb{P}\left(\bigcap_{n=1}^{\infty}A_n\right) =\lim_{n\to\infty}\left(\frac12+\frac{1}{2n}\right)=\frac12. $$ For \(n\ge1\), the portion removed in passing from \(A_n\) to \(A_{n+1}\) is \(A_n\setminus A_{n+1}\). Since the sequence is decreasing, \(A_{n+1}\subseteq A_n\), so finite additivity yields $$ \mathbb{P}(A_n\setminus A_{n+1}) =\mathbb{P}(A_n)-\mathbb{P}(A_{n+1}) =\left(\frac12+\frac{1}{2n}\right)-\left(\frac12+\frac{1}{2(n+1)}\right) =\frac{1}{2n(n+1)}. $$ These removed portions are pairwise disjoint. The total probability removed after \(A_1\), through the first \(N\) transitions, is $$ \sum_{n=1}^{N}\frac{1}{2n(n+1)} =\frac12\sum_{n=1}^{N}\left(\frac1n-\frac1{n+1}\right) =\frac12\left(1-\frac{1}{N+1}\right). $$ Because \(\mathbb{P}(A_1)=\frac12+\frac12=1\), the probability remaining in \(A_{N+1}\) is \(1-\frac12(1-\frac{1}{N+1})=\frac12+\frac{1}{2(N+1)}\), exactly the specified probability. As \(N\) tends to infinity, the removed probability tends to \(1/2\) and the remaining probability tends to \(1/2\), confirming the intersection calculation.
What Fails Without the Hypotheses?
Non-nested events do not satisfy either continuity formula just because their probabilities converge. For a concrete check, suppose a probability space has an event \(B\) with \(\mathbb{P}(B)=1/2\), and define \(E_n=B\) for odd \(n\) and \(E_n=B^c\) for even \(n\). Every term has probability \(1/2\), so \(\lim_n\mathbb{P}(E_n)=1/2\). But every tail contains both \(B\) and \(B^c\), so its union is \(\Omega\) and its intersection is \(\varnothing\). Consequently, $$ \bigcap_{N=1}^{\infty}\bigcup_{n\ge N}E_n=\Omega, \qquad \bigcup_{N=1}^{\infty}\bigcap_{n\ge N}E_n=\varnothing. $$ The first set contains the outcomes that occur in infinitely many of the events; the second contains those that occur in all sufficiently late events. Their probabilities are \(1\) and \(0\), respectively, not \(1/2\). This is not a counterexample to either theorem: the sequence is neither increasing nor decreasing.
Do not apply continuity from above to an arbitrary measure space without checking finiteness. For example, take \(\mathbb{N}=\{1,2,\dots\}\) with counting measure \(\mu\), and let \(A_n=\{n,n+1,\dots\}\). These events decrease and their intersection is empty, whose counting measure is \(0\). However, every \(A_n\) is infinite, so \(\mu(A_n)=\infty\) for every \(n\). Thus \(\lim_n\mu(A_n)=\infty\ne0=\mu(\bigcap_n A_n)\). Counting measure is not a probability measure and the sets do not have finite measure. For probability measures, this obstruction is ruled out by \(\mathbb{P}(\Omega)=1\).
In applications, the practical order is to check nesting first, identify the exact union or intersection, and only then pass to the limit of the probabilities. For increasing events, disjoint differences make the union probability an accumulated sum. For decreasing events, complements convert the intersection into an increasing union; finite total probability makes the subtraction valid. Neither argument asserts continuity for an arbitrary, non-nested event sequence.
Worked Example: A Decreasing Sequence Whose Intersection Is Empty
Suppose \((C_n)\) is decreasing and \(\mathbb{P}(C_n)=2^{-n}\) for \(n\ge1\). Determine the probability of its intersection and check the limit directly.
The sequence of probabilities is \(2^{-1},2^{-2},2^{-3},\dots\), which decreases to \(0\). Continuity from above therefore gives $$ \mathbb{P}\left(\bigcap_{n=1}^{\infty}C_n\right) =\lim_{n\to\infty}2^{-n}=0. $$ This also follows by examining the portion removed at each step: $$ \mathbb{P}(C_n\setminus C_{n+1}) =2^{-n}-2^{-(n+1)} =2^{-(n+1)}. $$ The removed portions for \(n=1,\dots,N\) are disjoint and have total probability $$ \sum_{n=1}^{N}2^{-(n+1)} =\frac12\left(1-2^{-N}\right). $$ Since \(\mathbb{P}(C_1)=2^{-1}=1/2\), the probability left in \(C_{N+1}\) is \(\frac12-\frac12(1-2^{-N})=2^{-(N+1)}\), as required. This remaining probability tends to \(0\), in agreement with the intersection calculation.
Check Your Understanding
For each question, first check whether the sequence is nested and whether its limit is a union or an intersection.
- If \((E_n)\) is increasing with \(\mathbb{P}(E_n)=1-3^{-n}\) for \(n\ge1\), what is \(\mathbb{P}(\bigcup_n E_n)\)?
- If \((A_n)\) is decreasing with \(\mathbb{P}(A_n)=\frac13+\frac{1}{3n}\) for \(n\ge1\), what is \(\mathbb{P}(\bigcap_n A_n)\)?
- In the proof of continuity from below, why are the sets \(D_n=E_n\setminus E_{n-1}\) pairwise disjoint?
- Give an example of a decreasing sequence of sets in a measure space where the measures of the sets do not converge to the measure of their intersection, and identify the missing hypothesis.
- For an event sequence that is not nested, does convergence of \(\mathbb{P}(E_n)\) alone determine the probability of its set limit? Explain.