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In probability theory, the Borel–Cantelli lemma is a theorem about sequences of events. In general, it is a result in measure theory. It is named after Émile Borel and Francesco Paolo Cantelli, who gave statement to the lemma in the first decades of the 20th century.^[1][2] A related result, sometimes called the second Borel–Cantelli lemma, is a partial converse of the first Borel–Cantelli lemma. The lemma states that, under certain conditions, an event will have probability of either zero or one. Accordingly, it is the best-known of a class of similar theorems, known as zero-one laws. Other examples include Kolmogorov's zero–one law and the Hewitt–Savage zero–one law.

Let E₁, E₂, ... be a sequence of events in some probability space. The Borel–Cantelli lemma states:^[3][4]

Here, "lim sup" denotes limit supremum of the sequence of events, and each event is a set of outcomes. That is, lim sup E_n is the set of outcomes that occur infinitely many times within the infinite sequence of events (E_n). Explicitly, $${\displaystyle \limsup _{n\to \infty }E_{n}=\bigcap _{n=1}^{\infty }\bigcup _{k=n}^{\infty }E_{k}.}$$The set lim sup E_n is sometimes denoted {E_n i.o.}, where "i.o." stands for "infinitely often". The theorem therefore asserts that if the sum of the probabilities of the events E_n is finite, then the set of all outcomes that are "repeated" infinitely many times must occur with probability zero. Note that no assumption of independence is required.

Suppose (X_n) is a sequence of random variables with Pr(X_n = 0) = 1/n² for each n. The probability that X_n = 0 occurs for infinitely many n is equivalent to the probability of the intersection of infinitely many [X_n = 0] events. The intersection of infinitely many such events is a set of outcomes common to all of them. However, the sum ΣPr(X_n = 0) converges to π²/6 ≈ 1.645 < ∞, and so the Borel–Cantelli Lemma states that the set of outcomes that are common to infinitely many such events occurs with probability zero. Hence, the probability of X_n = 0 occurring for infinitely many n is 0. Almost surely (i.e., with probability 1), X_n is nonzero for all but finitely many n.

Let (E_n) be a sequence of events in some probability space.

The sequence of events ${\textstyle \left\{\bigcup _{n=N}^{\infty }E_{n}\right\}_{N=1}^{\infty }}$ is non-increasing: $${\displaystyle \bigcup _{n=1}^{\infty }E_{n}\supseteq \bigcup _{n=2}^{\infty }E_{n}\supseteq \cdots \supseteq \bigcup _{n=N}^{\infty }E_{n}\supseteq \bigcup _{n=N+1}^{\infty }E_{n}\supseteq \cdots \supseteq \limsup _{n\to \infty }E_{n}.}$$By continuity from above, $${\displaystyle \Pr(\limsup _{n\to \infty }E_{n})=\lim _{N\to \infty }\Pr \left(\bigcup _{n=N}^{\infty }E_{n}\right).}$$By subadditivity, $${\displaystyle \Pr \left(\bigcup _{n=N}^{\infty }E_{n}\right)\leq \sum _{n=N}^{\infty }\Pr(E_{n}).}$$By original assumption, ${\textstyle \sum _{n=1}^{\infty }\Pr(E_{n})<\infty .}$ As the series ${\textstyle \sum _{n=1}^{\infty }\Pr(E_{n})}$ converges, $${\displaystyle \lim _{N\to \infty }\sum _{n=N}^{\infty }\Pr(E_{n})=0,}$$ as required.^[5]

For general measure spaces, the Borel–Cantelli lemma takes the following form:

A related result, sometimes called the second Borel–Cantelli lemma, is a partial converse of the first Borel–Cantelli lemma. The lemma states: If the events E_n are independent and the sum of the probabilities of the E_n diverges to infinity, then the probability that infinitely many of them occur is 1. That is:^[4]

The assumption of independence can be weakened to pairwise independence, but in that case the proof is more difficult.

The infinite monkey theorem follows from this second lemma.

The lemma can be applied to give a covering theorem in Rⁿ. Specifically (Stein 1993, Lemma X.2.1), if E_j is a collection of Lebesgue measurable subsets of a compact set in Rⁿ such that $${\displaystyle \sum _{j}\mu (E_{j})=\infty ,}$$ then there is a sequence F_j of translates $${\displaystyle F_{j}=E_{j}+x_{j}}$$ such that $${\displaystyle \lim \sup F_{j}=\bigcap _{n=1}^{\infty }\bigcup _{k=n}^{\infty }F_{k}=\mathbb {R} ^{n}}$$ apart from a set of measure zero.

Suppose that ${\textstyle \sum _{n=1}^{\infty }\Pr(E_{n})=\infty }$ and the events ${\displaystyle (E_{n})_{n=1}^{\infty }}$ are independent. It is sufficient to show the event that the E_n's did not occur for infinitely many values of n has probability 0. This is just to say that it is sufficient to show that $${\displaystyle 1-\Pr(\limsup _{n\to \infty }E_{n})=0.}$$

Noting that: $${\displaystyle {\begin{aligned}1-\Pr(\limsup _{n\to \infty }E_{n})&=1-\Pr \left(\{E_{n}{\text{ i.o.}}\}\right)=\Pr \left(\{E_{n}{\text{ i.o.}}\}^{c}\right)\\&=\Pr \left(\left(\bigcap _{N=1}^{\infty }\bigcup _{n=N}^{\infty }E_{n}\right)^{c}\right)=\Pr \left(\bigcup _{N=1}^{\infty }\bigcap _{n=N}^{\infty }E_{n}^{c}\right)\\&=\Pr \left(\liminf _{n\to \infty }E_{n}^{c}\right)=\lim _{N\to \infty }\Pr \left(\bigcap _{n=N}^{\infty }E_{n}^{c}\right),\end{aligned}}}$$ it is enough to show: ${\textstyle \Pr \left(\bigcap _{n=N}^{\infty }E_{n}^{c}\right)=0}$. Since the ${\displaystyle (E_{n})_{n=1}^{\infty }}$ are independent: $${\displaystyle {\begin{aligned}\Pr \left(\bigcap _{n=N}^{\infty }E_{n}^{c}\right)&=\prod _{n=N}^{\infty }\Pr(E_{n}^{c})\\&=\prod _{n=N}^{\infty }(1-\Pr(E_{n})).\end{aligned}}}$$ The convergence test for infinite products guarantees that the product above is 0, if ${\textstyle \sum _{n=N}^{\infty }\Pr(E_{n})}$ diverges. This completes the proof.

Another related result is the so-called counterpart of the Borel–Cantelli lemma. It is a counterpart of the Lemma in the sense that it gives a necessary and sufficient condition for the limsup to be 1 by replacing the independence assumption by the completely different assumption that ${\displaystyle (A_{n})}$ is monotone increasing for sufficiently large indices. This Lemma says:

Let ${\displaystyle (A_{n})}$ be such that ${\displaystyle A_{k}\subseteq A_{k+1}}$, and let ${\displaystyle {\bar {A}}}$ denote the complement of ${\displaystyle A}$. Then the probability of infinitely many ${\displaystyle A_{k}}$ occur (that is, at least one ${\displaystyle A_{k}}$ occurs) is one if and only if there exists a strictly increasing sequence of positive integers ${\displaystyle (t_{k})}$ such that $${\displaystyle \sum _{k}\Pr(A_{t_{k+1}}\mid {\bar {A}}_{t_{k}})=\infty .}$$This simple result can be useful in problems such as for instance those involving hitting probabilities for stochastic process with the choice of the sequence ${\displaystyle (t_{k})}$ usually being the essence.

Let ${\displaystyle (A_{n})}$ be a sequence of events with ${\textstyle \sum \Pr(A_{n})=\infty }$ and ${\textstyle \limsup _{k\to \infty }{\frac {\left(\sum _{n=1}^{k}\Pr(A_{n})\right)^{2}}{\sum _{1\leq m,n\leq k}\Pr(A_{m}\cap A_{n})}}>0.}$ Then there is a positive probability that ${\displaystyle A_{n}}$ occur infinitely often.

Let ${\displaystyle S_{m,n}=\sum _{i=m}^{n}\mathbf {1} _{A_{i}}}$. Then, note that $${\displaystyle E[S_{m,n}]^{2}=\left(\sum _{i=m}^{n}\Pr(A_{i})\right)^{2}}$$ and $${\displaystyle E[S_{m,n}^{2}]=\sum _{1\leq i\leq j\leq n}\Pr(A_{i}\cap A_{j}).}$$ Hence, we know that ${\displaystyle \limsup _{n\to \infty }{\frac {\mathbb {E} [S_{1,n}]^{2}}{\mathbb {E} [S_{1,n}^{2}]}}>0.}$ We have that $${\displaystyle \Pr \left(\bigcup _{i=m}^{n}A_{i}\right)=\Pr(S_{m,n}>0).}$$ Now, notice that by the Cauchy-Schwarz Inequality, $${\displaystyle \mathbb {E} [X]^{2}\leq \mathbb {E} [X\mathbf {1} _{\{X>0\}}]^{2}\leq \mathbb {E} [X^{2}]\Pr(X>0),}$$ therefore, $${\displaystyle \Pr(S_{m,n}>0)\geq {\frac {\mathbb {E} [S_{m,n}]^{2}}{\mathbb {E} [S_{m,n}^{2}]}}.}$$ We then have $${\displaystyle {\frac {\mathbb {E} [S_{m,n}]^{2}}{\mathbb {E} [S_{m,n}^{2}]}}\geq {\frac {E[S_{1,n}-S_{1,m-1}]^{2}}{E[S_{1,n}^{2}]}}.}$$ Given ${\displaystyle m}$, since ${\displaystyle \lim _{n\to \infty }\mathbb {E} [S_{1,n}]=\infty }$, we can find ${\displaystyle n}$ large enough so that $${\displaystyle {\biggr |}{\frac {\mathbb {E} [S_{1,n}]-\mathbb {E} [S_{1,m-1}]}{\mathbb {E} [S_{1,n}]}}-1{\biggr |}<\epsilon ,}$$ for any given ${\displaystyle \epsilon >0}$. Therefore, $${\displaystyle \lim _{m\to \infty }\sup _{n\geq m}\Pr \left(\bigcup _{i=m}^{n}A_{i}\right)\geq \lim _{m\to \infty }\sup _{n\geq m}{\frac {E[S_{1,n}]^{2}}{E[S_{1,n}^{2}]}}>0.}$$ But the left side is precisely the probability that the ${\displaystyle A_{n}}$ occur infinitely often since ${\displaystyle \{A_{k}{\text{ i.o.}}\}=\{\omega \in \Omega :\forall m,\exists n\geq m{\text{ s.t. }}\omega \in A_{n}\}.}$ We're done now, since we've shown that ${\displaystyle P(A_{k}{\text{ i.o.}})>0.}$

• Lévy's zero–one law
• Kuratowski convergence
• Infinite monkey theorem

• Prokhorov, A.V. (2001) [1994], "Borel–Cantelli lemma", Encyclopedia of Mathematics, EMS Press
• Feller, William (1961), An Introduction to Probability Theory and Its Application, John Wiley & Sons.
• Stein, Elias (1993), Harmonic analysis: Real-variable methods, orthogonality, and oscillatory integrals, Princeton University Press.
• Bruss, F. Thomas (1980), "A counterpart of the Borel Cantelli Lemma", J. Appl. Probab., 17 (4): 1094–1101, doi:10.2307/3213220, JSTOR 3213220, S2CID 250344204.
• Durrett, Rick. "Probability: Theory and Examples." Duxbury advanced series, Third Edition, Thomson Brooks/Cole, 2005.

• Planet Math Proof Refer for a simple proof of the Borel Cantelli Lemma