Sharpness of the Borel–Cantelli Hypotheses

Why the Probability Sum Is Not the Whole Story

For a sequence of events, the series of their probabilities is a useful measure of how much probability mass the sequence carries. But that series does not, by itself, describe how occurrences are arranged together. Events may overlap heavily, or their occurrences may be independent.

The Borel–Cantelli lemmas make this distinction precise. If the probability sum is finite, only finitely many events occur almost surely, regardless of dependence. If the sum diverges, the conclusion can be very different: independence is a sufficient condition for infinitely many occurrences almost surely, but divergence alone is not.

Definition: A sequence of events \(A_1,A_2,\ldots\) occurs infinitely often if, for every \(m\in\mathbb{N}\), at least one event \(A_n\) with \(n\ge m\) occurs. The event that this happens is denoted by

\{A_n\ \text{infinitely often}\} = \bigcap_{m=1}^{\infty}\bigcup_{n=m}^{\infty}A_n.

This definition captures the idea that no matter how far along the sequence one starts, another occurrence remains. Its complement is the event that only finitely many \(A_n\) occur.

The First Borel–Cantelli Lemma

The first lemma requires no independence assumption. Its hypothesis is that the total sum of event probabilities is finite.

Theorem (First Borel–Cantelli Lemma): If \((A_n)_{n\in\mathbb{N}}\) is a sequence of events and

\sum_{n=1}^{\infty}\mathbb{P}(A_n)<\infty,

then \(\mathbb{P}(A_n\text{ infinitely often})=0\).

Proof. For every \(m\), the event that \(A_n\) occurs infinitely often is contained in \(\bigcup_{n=m}^{\infty}A_n\). Countable subadditivity therefore gives

\mathbb{P}(A_n\text{ infinitely often}) \leq \mathbb{P}\!\left(\bigcup_{n=m}^{\infty}A_n\right) \leq \sum_{n=m}^{\infty}\mathbb{P}(A_n).

Since the full series is finite, its tails tend to zero as \(m\) tends to infinity. The left-hand side does not depend on \(m\), so it must be zero. \(\square\)

The mechanism is simple: a finite total probability leaves arbitrarily little probability in the far tail. Infinitely many occurrences would require an occurrence in every tail, and the probability of that possibility is bounded by each tail sum. No assumption about how the events depend on one another entered the argument.

What Divergence Does—and Does Not—Imply

The converse is not true in general. A divergent sum of probabilities does not guarantee that events occur infinitely often. To see why, consider a sequence in which later events are contained in earlier ones. Their probabilities can add up to infinity even though, outside a null set, a point belongs to only finitely many of them.

Worked Example: Nested Events with a Divergent Sum

Take \(\Omega=[0,1]\) with Lebesgue probability and define \(A_n=[0,1/n]\) for every \(n\in\mathbb{N}\). Then \(A_1=[0,1]\), and for every \(n\),

\mathbb{P}(A_n)=\frac{1}{n}.

The harmonic series diverges. For example, the terms from \(2^k\) through \(2^{k+1}-1\) have sum at least \(2^k/2^{k+1}=1/2\), for every \(k\in\mathbb{N}_0\). Thus \(\sum_n\mathbb{P}(A_n)=\infty\).

Nevertheless, if \(x>0\), then \(1/n<x\) for all sufficiently large \(n\). For those \(n\), \(x\notin[0,1/n]\), so \(x\) belongs to only finitely many \(A_n\). The point \(x=0\) belongs to every \(A_n\). Consequently,

\{A_n\text{ infinitely often}\}=\{0\}, \qquad \mathbb{P}(A_n\text{ infinitely often})=0.

The events are nested, so their occurrences are strongly dependent. The divergent sum alone does not prevent the probability mass from being concentrated on a shrinking collection of outcomes.

This example shows the sharp difference between the two directions. Summability is enough for the first lemma, without any dependence condition. Divergence is not enough for the reverse conclusion; some control of dependence is needed for a general theorem guaranteeing infinitely many occurrences.

The Second Borel–Cantelli Lemma

A standard such control is mutual independence. A sequence of events is mutually independent when every finite subcollection is independent: for distinct indices \(i_1,\ldots,i_k\), the probability of their intersection is the product of their probabilities.

Theorem (Second Borel–Cantelli Lemma): Let \((A_n)_{n\in\mathbb{N}}\) be a mutually independent sequence of events. If

\sum_{n=1}^{\infty}\mathbb{P}(A_n)=\infty,

then \(\mathbb{P}(A_n\text{ infinitely often})=1\).

Proof. Fix \(m\in\mathbb{N}\). For \(N\ge m\), let \(F_{m,N}=\bigcap_{n=m}^{N}A_n^c\), the event that none of \(A_m,\ldots,A_N\) occurs. By mutual independence,

\mathbb{P}(F_{m,N}) = \prod_{n=m}^{N}\bigl(1-\mathbb{P}(A_n)\bigr) \leq \exp\!\left(-\sum_{n=m}^{N}\mathbb{P}(A_n)\right).

Here the inequality follows from \(1-x\leq e^{-x}\) for \(0\leq x\leq1\). Because the probability series diverges, its tail from \(m\) also diverges. The right-hand side therefore tends to zero as \(N\) tends to infinity. The events \(F_{m,N}\) decrease to \(\bigcap_{n=m}^{\infty}A_n^c\); by continuity of probability for decreasing events,

\mathbb{P}\!\left(\bigcap_{n=m}^{\infty}A_n^c\right)=0.

The event that only finitely many \(A_n\) occur is \(\bigcup_{m=1}^{\infty}\bigcap_{n=m}^{\infty}A_n^c\). It is a countable union of null events and hence has probability zero. Its complement, the event that \(A_n\) occurs infinitely often, has probability one. \(\square\)

The proof identifies the role of independence: it makes the probability of avoiding every event in a finite block equal to a product. The inequality \(1-x\leq e^{-x}\) then bounds that product by an exponential of the negative probability sum. When the sum diverges, the probability of avoiding all events in any fixed tail is zero.

Dependence Can Change the Conclusion

A second dependent example isolates another point. Even when every event has the same positive probability, infinitely many occurrences need not have probability zero or one. The joint arrangement of events matters, not just their individual probabilities.

Worked Example: Repeating One Event

Let \(\Omega=[0,1]\) with Lebesgue probability, let \(A=[0,1/2]\), and set \(A_n=A\) for every \(n\in\mathbb{N}\). For each \(n\), \(\mathbb{P}(A_n)=1/2\), so

\sum_{n=1}^{\infty}\mathbb{P}(A_n) = \sum_{n=1}^{\infty}\frac12 = \infty.

If \(x\in A\), then \(x\) belongs to every \(A_n\), and thus infinitely many occur. If \(x\notin A\), then \(x\) belongs to none of them. Hence

\{A_n\text{ infinitely often}\}=A, \qquad \mathbb{P}(A_n\text{ infinitely often})=\frac12.

The events are maximally dependent: they are identical. The divergent probability sum does not force the infinitely-often probability to equal one.

Mutual independence is therefore a useful sufficient hypothesis for the second lemma, not a claim that every dependent sequence must fail its conclusion. Some dependent sequences do have infinitely many occurrences almost surely. The point is that divergence alone gives no universal guarantee.

Worked Example: Independent Events with the Same Tail Probabilities

Let \(\Omega=\{0,1\}^{\{2,3,\ldots\}}\) carry the product probability measure under which coordinate \(n\) equals \(1\) with probability \(1/n\) and equals \(0\) with probability \(1-1/n\). Let \(B_n\) be the event that coordinate \(n\) equals \(1\), for \(n\ge2\), and set \(B_1=\varnothing\). The coordinate events are mutually independent, and \(B_1\) is independent of them as well.

For \(n\ge2\), \(\mathbb{P}(B_n)=1/n\). Thus the sum of the probabilities diverges, since the harmonic series from \(n=2\) onward diverges. The second Borel–Cantelli lemma gives

\mathbb{P}(B_n\text{ infinitely often})=1.

Compare this with the nested example: the probabilities agree for every \(n\ge2\), but the first probabilities differ. In the nested example, \(\mathbb{P}(A_1)=1\); here, \(\mathbb{P}(B_1)=0\). This discrepancy does not affect divergence of either sum or the infinitely-often conclusion. The contrast is that the independent events occur infinitely often almost surely, whereas the nested events do so only on a null set.

Reading the Hypotheses Carefully

The Borel–Cantelli lemmas are often remembered as a pair, but their assumptions should not be merged. The first lemma says that summability rules out infinitely many occurrences almost surely, and it places no independence requirement on the events. The second lemma says that divergence guarantees infinitely many occurrences almost surely under mutual independence.

Probability sum	Dependence condition	Conclusion
\(\sum_n\mathbb{P}(A_n)<\infty\)	None required	Only finitely many \(A_n\) occur almost surely
\(\sum_n\mathbb{P}(A_n)=\infty\)	Mutual independence	Infinitely many \(A_n\) occur almost surely
\(\sum_n\mathbb{P}(A_n)=\infty\)	No dependence control assumed	No general infinitely-often conclusion

A common mistake is to infer from the first lemma that divergence must imply infinitely many occurrences. The nested-interval construction disproves that inference. Another is to treat independence as necessary for every individual sequence to have infinitely many occurrences almost surely. It is not: the theorem gives a broadly useful sufficient condition, while particular dependent sequences may still satisfy the same conclusion for other reasons.

The essential lesson is that marginal probabilities alone do not determine the behavior of an event sequence. A probability sum records the sizes of individual events, but dependence governs how those events fit together. For a divergent sum, that distinction can decide whether infinitely many occurrences have probability zero, one-half, or one.

Check Your Understanding

Try to answer each question by identifying both the probability-sum condition and the dependence information involved.

Why does the first Borel–Cantelli lemma not need independence?
For the events \(A_n=[0,1/n]\) on \([0,1]\), identify the infinitely-often event and calculate its probability.
In the proof of the second Borel–Cantelli lemma, where is mutual independence used?
For the repeated-event example, what is the probability sum, and what is the probability of infinitely many occurrences?
Why does the independent sequence with \(B_1=\varnothing\) have matching probabilities with the nested sequence only for \(n\ge2\), and why does that difference not affect the comparison?