Proof of Borel-Cantelli Lemmas

Write $p_{k}=P(A_{k})$, a real number in $[0,1]$, $A_{k}^{c}=\Omega\setminus A_{k}$ and $L=\{A_{k}\text{ infinitely often}\}$. Throughout, complements, finite and countable unions and intersections of events are events (Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (x), Unions and Intersections over a Countable Index Set in a σ-Algebra (i), Unions and Intersections over a Countable Index Set in a σ-Algebra (ii)), a finite sum of the $p_{k}$ formed in $[0,\infty]$ is their real sum (A Finite Sum of Nonnegative Real Numbers in [0,∞] Is Their Real Sum (i)), and for $n\le N$ in $\mathbb{N}$, $\bigcap_{k=n}^{N}$, $\prod_{k=n}^{N}$ and $\sum_{k=n}^{N}$ stand for the intersection, product and sum over $j\in[N-n+1]$ of the terms with index $k=n-1+j$.

Proof of (i). As $\sum_{k\in\mathbb{N}}p_{k}$ is real, the series $\sum_{k=1}^{\infty}p_{k}$ converges and has this sum (Sums of Sequences in [0,∞]: Real Series, Linearity, Comparison, Rearrangement, Double Series and Monotone Convergence (vi)), so its tails $t_{q}=\sum_{k=1}^{\infty}p_{q+k}$, $q\in\mathbb{N}$, form a null sequence (Elementary Properties of Series of Real Numbers: Linearity, Null Terms, the Cauchy Criterion, Index Shifts, Tails, Order and Telescoping (v)), and $t_{q}=\sum_{k\in\mathbb{N}}p_{q+k}$, again by Sums of Sequences in [0,∞]: Real Series, Linearity, Comparison, Rearrangement, Double Series and Monotone Convergence (vi). Let $q\in\mathbb{N}$. As the $k\ge q+1$ are exactly the $q+k'$ with $k'\in\mathbb{N}$, $\bigcup_{k\ge q+1}A_{k}=\bigcup_{k'\in\mathbb{N}}A_{q+k'}$, so monotonicity (Basic Properties of a Measure (iii)) and countable subadditivity give

$$0\le P(L)\le P\Bigl(\bigcup_{k\ge q+1}A_{k}\Bigr)\le\sum_{k'\in\mathbb{N}}p_{q+k'}=t_{q}.$$

As $t_{q}\to0$, $P(L)\le0$ (Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (x)), so $P(L)=0$.

Proof of (ii). Independence of the complements. Each $\{A_{k}\}$ is a $\pi$-system, so the family $(\sigma(\{A_{k}\}))_{k\in\mathbb{N}}$ is independent (Independence of Events and of Random Elements (i), Independence of σ-Algebras: π-Systems, Grouping and Pairs (ii)), and $A_{k}^{c}\in\sigma(\{A_{k}\})$ (Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (iii)). Hence for $n\le N$ in $\mathbb{N}$ the product formula (Independence of Classes of Events and of σ-Algebras (i)) for the distinct indices $n,\dots,N$ and the complement rule (Basic Properties of a Measure (v)) give

$$P\Bigl(\bigcap_{k=n}^{N}A_{k}^{c}\Bigr)=\prod_{k=n}^{N}(1-p_{k})\le\prod_{k=n}^{N}\exp(-p_{k})=\exp\Bigl(-\sum_{k=n}^{N}p_{k}\Bigr),$$

the inequality and the last equality by induction on $N$ from $0\le1-p_{k}\le\exp(-p_{k})$ (Basic Properties of the Exponential Function (v)) and $\exp(u+v)=\exp(u)\exp(v)$ (Basic Properties of the Exponential Function (ii)).

Each tail union has probability one. Fix $n\in\mathbb{N}$ and let $D_{n}=\bigcap_{k\ge n}A_{k}^{c}=\Omega\setminus\bigcup_{k\ge n}A_{k}$. Let $\varepsilon>0$, and choose a real $M\ge0$ with $\exp(u)<\varepsilon$ for all $u<-M$ (Basic Properties of the Exponential Function (x), enlarging $M$ if necessary). With $s_{N}=\sum_{k=1}^{N}p_{k}$, we have $\sup_{N}s_{N}=\sum_{k\in\mathbb{N}}p_{k}=\infty$ (Sum of a Sequence in [0,∞] (ii)), so there is $N\in\mathbb{N}$ with $s_{N}>M+s_{n}$ (Order and Suprema in the Extended Real Line (x)); then $N>n$, as $(s_{N})$ is nondecreasing (Sums of Sequences in [0,∞]: Real Series, Linearity, Comparison, Rearrangement, Double Series and Monotone Convergence (i)), and $\sum_{k=n}^{N}p_{k}\ge s_{N}-s_{n}>M$ (Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms (iii)). Since $D_{n}\subseteq\bigcap_{k=n}^{N}A_{k}^{c}$, monotonicity and the previous step give

$$P(D_{n})\le\exp\Bigl(-\sum_{k=n}^{N}p_{k}\Bigr)<\varepsilon.$$

As $\varepsilon>0$ was arbitrary, $P(D_{n})=0$, so $P\bigl(\bigcup_{k\ge n}A_{k}\bigr)=1$ by the complement rule.

Intersection over $n$. The events $\bigcup_{k\ge n}A_{k}$ decrease in $n$ and all have probability $1$, so by continuity from above (Basic Properties of a Measure (ix))

$$P(L)=P\Bigl(\bigcap_{n\in\mathbb{N}}\ \bigcup_{k\ge n}A_{k}\Bigr)=\inf_{n}1=1.$$