From a Divergent Sum to Infinitely Many Occurrences
For events \(A_1,A_2,\ldots\) in a probability space, “\(A_n\) occurs infinitely often” means that every tail of the sequence contains at least one occurrence. The corresponding event is the set-theoretic limsup, \[ \limsup_{n\to\infty} A_n=\bigcap_{N=1}^{\infty}\bigcup_{n\ge N}A_n. \] The first Borel-Cantelli lemma gives a sufficient condition for this event to have probability zero: if the sum of the event probabilities is finite, only finitely many events occur almost surely. The second lemma addresses the opposite direction, but requires an additional hypothesis. If the events are mutually independent and their probabilities have divergent sum, then infinitely many occur almost surely.
Throughout, \((\Omega,\mathcal{F},\mathbb{P})\) is a probability space, so \(\mathbb{P}(\Omega)=1\); in particular, it is finite. The events \(A_n\) belong to \(\mathcal{F}\), and \(p_n=\mathbb{P}(A_n)\in[0,1]\). No completeness assumption is needed. We use continuity of probability: for an increasing sequence of events \(E_j\), \(\mathbb{P}(\bigcup_jE_j)=\lim_j\mathbb{P}(E_j)\); for a decreasing sequence, \(\mathbb{P}(\bigcap_jE_j)=\lim_j\mathbb{P}(E_j)\). These follow from countable additivity and finiteness of the probability measure.
The Second Borel-Cantelli Lemma
Proof. Fix \(N\ge1\). For \(M\ge N\), let \(C_{N,M}=\bigcap_{n=N}^{M}A_n^c\), the event that none of \(A_N,\ldots,A_M\) occurs. Mutual independence implies \[ \mathbb{P}(C_{N,M})=\prod_{n=N}^{M}(1-p_n). \] To justify independence of the complements from the stated definition, expand the finite intersection using inclusion-exclusion. Each intersection of selected events has probability equal to the product of its \(p_n\)'s; summing with alternating signs gives exactly \(\prod_{n=N}^{M}(1-p_n)\). This argument includes \(p_n=0\) and \(p_n=1\).
As \(M\) increases, \(C_{N,M}\) decreases, and its intersection over \(M\ge N\) is \(C_N=\bigcap_{n\ge N}A_n^c\), the event that no event from index \(N\) onward occurs. Continuity from above applies because \(\mathbb{P}(C_{N,N})\le1<\infty\), so \[ \mathbb{P}(C_N)=\lim_{M\to\infty}\prod_{n=N}^{M}(1-p_n). \] For \(0\le x\le1\), \(1-x\le e^{-x}\). For example, on \(0\le x<1\), this follows from \(\ln(1-x)\le-x\), and at \(x=1\) it holds directly. Therefore \[ 0\le\prod_{n=N}^{M}(1-p_n)\le \exp\left(-\sum_{n=N}^{M}p_n\right). \] The series \(\sum_np_n\) diverges, so every tail \(\sum_{n=N}^{\infty}p_n\) diverges: removing finitely many finite terms cannot turn an infinite sum into a finite one. Hence the upper bound tends to zero as \(M\to\infty\), and \(\mathbb{P}(C_N)=0\).
The event that only finitely many \(A_n\) occur is \(\bigcup_{N=1}^{\infty}C_N\): if only finitely many occur, some tail contains none, and if a tail contains none, only finitely many can have occurred. Countable subadditivity now gives \(\mathbb{P}(\bigcup_N C_N)\le\sum_N\mathbb{P}(C_N)=0\). Its complement is \(\limsup_n A_n\), so that event has probability one. \(\square\)
Worked Applications
Worked Example: Probabilities \(1/(n+1)\)
Suppose the events \(A_n\) are mutually independent and \(\mathbb{P}(A_n)=1/(n+1)\) for \(n\ge1\). We verify divergence and calculate the probability of no occurrences on a finite block.
For \(N\le M\), independence gives \[ \mathbb{P}\left(\bigcap_{n=N}^{M}A_n^c\right) =\prod_{n=N}^{M}\left(1-\frac1{n+1}\right) =\prod_{n=N}^{M}\frac{n}{n+1} =\frac{N}{M+1}. \] The factors cancel consecutively: the numerator is \(N(N+1)\cdots M\), and the denominator is \((N+1)(N+2)\cdots(M+1)\). For fixed \(N\), this probability tends to \(0\) as \(M\to\infty\).
The probability sum diverges. Indeed, for each integer \(k\ge1\), the block \(2^k\le n\le2^{k+1}-1\) contains \(2^k\) terms, and for every such \(n\), \(1/(n+1)\ge1/2^{k+1}\). Its block sum is therefore at least \(2^k/2^{k+1}=1/2\). Infinitely many blocks contribute at least \(1/2\), so the full sum is infinite. The theorem gives \(\mathbb{P}(\limsup_n A_n)=1\). Check: taking \(N=2\) and \(M=5\), the four factors are \(2/3,3/4,4/5,5/6\), whose product is \(1/3\). The formula gives \(2/(5+1)=1/3\) as well.
Worked Example: Fair Independent Events
Let \((A_n)\) be mutually independent with \(\mathbb{P}(A_n)=1/2\) for every \(n\). For any \(N\le M\), there are \(M-N+1\) events in the block, and \[ \mathbb{P}\left(\bigcap_{n=N}^{M}A_n^c\right) =\prod_{n=N}^{M}\left(1-\frac12\right) =\left(\frac12\right)^{M-N+1}. \] For fixed \(N\), the exponent \(M-N+1\) tends to infinity, so this probability tends to zero. Also, every partial sum of the probabilities is \(\sum_{n=1}^{M}1/2=M/2\), which tends to infinity. Thus the second Borel-Cantelli lemma implies that infinitely many \(A_n\) occur with probability one. Check: for the block \(n=3,4,5\), the three complement probabilities multiply to \((1/2)^3=1/8\), agreeing with the formula using exponent \(5-3+1=3\).
Worked Example: A Finite Sum Gives the Opposite Conclusion
Suppose the events \(A_n\) are mutually independent and \(\mathbb{P}(A_n)=2^{-n}\) for \(n\ge1\). The geometric series has sum \[ \sum_{n=1}^{\infty}2^{-n}=\frac{1/2}{1-1/2}=1<\infty. \] The first Borel-Cantelli lemma says that \(\mathbb{P}(\limsup_n A_n)=0\); independence is not needed for this direction. In other words, only finitely many of these events occur almost surely.
For completeness, the finite-block probability of no occurrence is also explicit: \[ \mathbb{P}\left(\bigcap_{n=1}^{M}A_n^c\right) =\prod_{n=1}^{M}(1-2^{-n}). \] This product is not needed to apply the first lemma: the relevant exact quantity is the probability sum \(1\). Check: its first three terms are \(1/2+1/4+1/8=7/8\), and the remaining geometric tail is \((1/8)/(1-1/2)=1/4\); their sum is \(7/8+1/4=9/8\), so this purported check would be inconsistent. Correctly, the tail after \(n=3\) starts at \(2^{-4}=1/16\), and equals \((1/16)/(1-1/2)=1/8\). Thus the full sum is \(7/8+1/8=1\), as claimed.
Comparison and a Common Pitfall
The two Borel-Cantelli conclusions use opposite conditions and have different logical force. The first lemma applies to any events when the sum of their probabilities is finite. The second lemma applies when the events are mutually independent and the sum is infinite. Neither conclusion says that the events must occur consecutively or that any particular event has positive probability of occurring after a specified index.
| Probability sum | Additional condition | Conclusion about \(\limsup A_n\) |
|---|---|---|
| \(\sum_n\mathbb{P}(A_n)<\infty\) | None | Probability \(0\): only finitely many occur almost surely |
| \(\sum_n\mathbb{P}(A_n)=\infty\) | Mutual independence | Probability \(1\): infinitely many occur almost surely |
| \(\sum_n\mathbb{P}(A_n)=\infty\) | No independence assumption | No general conclusion from the sum alone |
Common mistake: using the second lemma without independence. Let \(B\) be one event in a probability space with \(\mathbb{P}(B)=1/2\), and set \(A_n=B\) for every \(n\). Then \(\mathbb{P}(A_n)=1/2\), so \(\sum_n\mathbb{P}(A_n)=\infty\). But \(\limsup_n A_n=B\), since either every \(A_n\) occurs or none does; consequently \(\mathbb{P}(\limsup_n A_n)=1/2\), not \(1\). The events are not independent: \(\mathbb{P}(A_1\cap A_2)=1/2\), whereas \(\mathbb{P}(A_1)\mathbb{P}(A_2)=1/4\). The divergent sum alone cannot replace the independence hypothesis.
The proof also indicates precisely where independence enters: it turns the probability of avoiding all events in a finite block into a product of complement probabilities. The exponential bound then forces the probability of avoiding an entire tail to be zero. Without that product identity, the argument does not establish that a tail has zero probability of being empty.
Check Your Understanding
For each question, identify whether the first or second Borel-Cantelli lemma applies and verify its hypotheses.
- For mutually independent events with \(\mathbb{P}(A_n)=1/(n+2)\), show that the probability sum diverges and state the almost-sure conclusion.
- For events with \(\mathbb{P}(A_n)=3^{-n}\), what can the first Borel-Cantelli lemma conclude, and is independence needed?
- In the proof of the second lemma, why does continuity from above apply to \(\bigcap_{M\ge N}C_{N,M}\)?
- Construct a sequence of events with divergent probability sum for which the probability of infinitely many occurrences is strictly between zero and one.
- Explain why pairwise independence is not the independence hypothesis stated in the second Borel-Cantelli lemma.