Limsup Events and Infinitely Often Occurrence

From Repeated Events to a Single Event

Given a sequence of events, we may want to know whether one of them occurs, whether at least one occurs after a specified time, or whether occurrences continue indefinitely. The last question is captured by one event built from the whole sequence: its limsup.

The limsup is useful because it turns an informal statement about infinitely many separate occurrences into a precise set operation. Once that event is identified, the Borel–Cantelli lemmas give conditions under which it has probability zero or one.

The Limsup of a Sequence of Events

Definition: Let $(E_n)_{n\in\mathbb N}$ be a sequence of events in a probability space $(\Omega,\mathcal F,\mathbb P)$. Its limsup is the event consisting of outcomes that belong to events $E_n$ for infinitely many indices $n$. In symbols, $$ \limsup_{n\to\infty} E_n = \bigcap_{N=1}^{\infty}\ \bigcup_{n\ge N} E_n. $$

For each $N$, the union $\bigcup_{n\ge N}E_n$ consists of outcomes for which at least one event occurs at an index $n\ge N$. Taking the intersection over all $N$ requires that this remain true no matter how far along the sequence we start. That is exactly the condition of infinitely many occurrences.

More explicitly, an outcome $\omega$ lies in the limsup if and only if, for every $N\in\mathbb N$, there exists $n\ge N$ such that $\omega\in E_n$. A finite collection of occurrences cannot satisfy this condition: after the last occurrence, one can choose $N$ larger than its index. Conversely, infinitely many occurrences include one beyond every $N$.

Because $\mathcal F$ is a sigma-algebra, it is closed under countable unions and intersections. Thus the limsup is an event whenever every $E_n$ is an event. The equivalent notation $\{E_n\ \text{infinitely often}\}$, often abbreviated $\{E_n\ \mathrm{i.o.}\}$, refers to this same event.

Worked Example: Nested Intervals and a Divergent Sum

Take $\Omega=[0,1]$ with its Borel sigma-algebra and uniform probability measure, and define $E_n=[0,1/n]$ for $n\ge 2$. For each $N\ge2$, the intervals are nested, so $$ \bigcup_{n\ge N} E_n=[0,1/N]. $$ Consequently, $$ \limsup_{n\to\infty}E_n = \bigcap_{N\ge2}[0,1/N] = \{0\}. $$ The only outcome in infinitely many of these intervals is $0$. Its probability is zero. At the same time, $$ \sum_{n=2}^{\infty}\mathbb P(E_n) = \sum_{n=2}^{\infty}\frac1n = \infty. $$ This does not contradict the second Borel–Cantelli lemma: the events are not independent. For example, $E_2\cap E_3=E_3$, so $\mathbb P(E_2\cap E_3)=1/3$, whereas $\mathbb P(E_2)\mathbb P(E_3)=1/6$.

The First Borel–Cantelli Lemma

Theorem (First Borel–Cantelli lemma): If $(E_n)_{n\in\mathbb N}$ is a sequence of events and $$ \sum_{n=1}^{\infty}\mathbb P(E_n)<\infty, $$ then $$ \mathbb P\!\left(\limsup_{n\to\infty}E_n\right)=0. $$ No independence assumption is required.

Proof. Fix $N\in\mathbb N$. By the definition of the limsup, $$ \limsup_{n\to\infty}E_n \subseteq \bigcup_{n\ge N}E_n. $$ Countable subadditivity therefore gives $$ \mathbb P\!\left(\limsup_{n\to\infty}E_n\right) \le \mathbb P\!\left(\bigcup_{n\ge N}E_n\right) \le \sum_{n\ge N}\mathbb P(E_n). $$ The series of probabilities converges, so its tail tends to zero as $N\to\infty$. The nonnegative quantity on the left is bounded above by every such tail, hence it must equal zero. $$ \square $$

The key feature is that summability makes the probability of any occurrence after a late index small. An outcome in the limsup must have an occurrence after every index, so its probability cannot exceed any of these vanishing tail bounds. This argument uses only countable subadditivity.

Worked Example: Disjoint Events with Summable Probabilities

Again use $[0,1]$ with uniform probability, and let $$ E_n=\left(\frac1{n+1},\frac1n\right],\qquad n\in\mathbb N. $$ The events are pairwise disjoint, and $$ \mathbb P(E_n)=\frac1n-\frac1{n+1}=\frac1{n(n+1)}. $$ The probabilities telescope: $$ \sum_{n=1}^{\infty}\mathbb P(E_n) = \sum_{n=1}^{\infty}\left(\frac1n-\frac1{n+1}\right) = 1. $$ Since an outcome belongs to at most one of these disjoint events, no outcome belongs to infinitely many of them. Thus $\limsup_n E_n=\varnothing$, and its probability is zero, as the first Borel–Cantelli lemma guarantees.

The Second Borel–Cantelli Lemma

Theorem (Second Borel–Cantelli lemma): Suppose the events $(E_n)_{n\in\mathbb N}$ are mutually independent and $$ \sum_{n=1}^{\infty}\mathbb P(E_n)=\infty. $$ Then $$ \mathbb P\!\left(\limsup_{n\to\infty}E_n\right)=1. $$ Here mutual independence means that every finite subfamily of the events is independent.

Proof. Write $p_n=\mathbb P(E_n)$, and for $N\in\mathbb N$ set $$ F_N=\bigcap_{n\ge N}E_n^c. $$ This is the event that none of the events $E_n$ occurs at or after $N$. For $M\ge N$, mutual independence gives $$ \mathbb P\!\left(\bigcap_{n=N}^{M}E_n^c\right) = \prod_{n=N}^{M}(1-p_n) \le \exp\!\left(-\sum_{n=N}^{M}p_n\right). $$ The inequality follows from $1-x\le e^{-x}$ for $x\in[0,1]$, applied to each $p_n$. Since the full series $\sum_n p_n$ diverges, its tail starting at $N$ also diverges. Therefore the right-hand side tends to zero as $M\to\infty$. The finite intersections $\bigcap_{n=N}^{M}E_n^c$ decrease to $F_N$; continuity of probability for decreasing sequences gives $\mathbb P(F_N)=0$.

An outcome fails to belong to the limsup exactly when it has only finitely many occurrences, which means there is some $N$ after which no $E_n$ occurs. Thus $$ \left(\limsup_{n\to\infty}E_n\right)^c = \bigcup_{N=1}^{\infty}F_N. $$ By countable subadditivity, this complement has probability at most $\sum_{N=1}^{\infty}\mathbb P(F_N)=0$. The limsup therefore has probability one. $$ \square $$

Independence is the ingredient that turns the probability of avoiding a finite block into a product. The divergent sum then forces the probability of avoiding longer and longer blocks to tend to zero. Pairwise independence alone is not the hypothesis stated here; the argument uses independence of every finite subfamily.

Worked Example: Infinitely Many Heads in Independent Coin Tosses

Let $\Omega=\{0,1\}^{\mathbb N}$ with the probability measure under which the coordinates are independent and each coordinate equals $1$ with probability $1/2$. Define $E_n=\{\omega:\omega_n=1\}$, the event that toss $n$ is heads. The events are mutually independent and $$ \sum_{n=1}^{\infty}\mathbb P(E_n) = \sum_{n=1}^{\infty}\frac12 = \infty. $$ The second Borel–Cantelli lemma applies, giving $$ \mathbb P(\text{heads occurs infinitely often}) = \mathbb P\!\left(\limsup_{n\to\infty}E_n\right) = 1. $$ This is an almost-sure statement, not a claim that every possible sequence of tosses has infinitely many heads. The all-tails sequence, for instance, has no heads, but it is a single outcome of probability zero.

How to Choose the Right Lemma

The two lemmas have different hypotheses and conclusions. The first is a one-way implication from a finite sum to probability zero and requires no independence. The second gives probability one from a divergent sum, but only under mutual independence. A divergent sum by itself is not enough, as the nested-interval example showed.

Condition on the probabilities	Additional assumption	Conclusion about the limsup
$\sum_n\mathbb P(E_n)<\infty$	None	$\mathbb P(\limsup_n E_n)=0$
$\sum_n\mathbb P(E_n)=\infty$	The events are mutually independent	$\mathbb P(\limsup_n E_n)=1$

A common error is to use the second lemma after checking only that the probabilities have a divergent sum. Always check the independence condition separately. Conversely, independence is unnecessary for the first lemma, so requiring it there would impose an irrelevant extra assumption.

A Reliable Application Strategy

When faced with a question about events occurring infinitely often, first translate the phrase into the limsup. Then check the sum of the event probabilities and, if that sum diverges, check whether the required independence is available.

Write the event.
Use $\limsup_n E_n=\bigcap_N\bigcup_{n\ge N}E_n$ to make “infinitely often” precise.

Inspect the probability series.
If its sum is finite, the first Borel–Cantelli lemma gives a null limsup.

Check independence when the sum diverges.
If the events are mutually independent, the second lemma gives a limsup of probability one. Without that hypothesis, the divergent sum alone does not settle the question.

These results distinguish two ways that infinitely many occurrences can be controlled. Summable probabilities rule them out almost surely even when events have strong dependence. A divergent sum predicts infinitely many occurrences almost surely only when independence prevents the events from clustering in an excessively dependent way.

Check Your Understanding

Use the set definition of the limsup and the hypotheses of the Borel–Cantelli lemmas to answer the following.

Write $\limsup_n E_n$ using countable unions and intersections, and explain its meaning for a single outcome.
Why does the first Borel–Cantelli lemma not require independence?
What precise independence assumption is used in the proof of the second Borel–Cantelli lemma?
For $E_n=[0,1/n]$ under uniform probability on $[0,1]$, find the limsup and explain why the divergent probability sum does not yield an almost-sure conclusion.
State the conclusion available when the event probabilities have a divergent sum but mutual independence has not been established.

Condition on the probabilities	Additional assumption	Conclusion about the limsup
\(\sum_n\mathbb P(E_n)<\infty\)	None	\(\mathbb P(\limsup_n E_n)=0\)
\(\sum_n\mathbb P(E_n)=\infty\)	The events are mutually independent	\(\mathbb P(\limsup_n E_n)=1\)