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Last updated on: Sun Jul 19 09:18:17 IST 2026


$\newcommand{\io}{~~\mathrm{i.o.}}$ $\newcommand{\ev}{~~\mathrm{ev.}}$ $\newcommand{\calF}{\mathcal{F}}$ $\newcommand{\rightarrowL}[1]{\stackrel{L_{#1}}{\longrightarrow}}$ $\newcommand{\rightarrowP}{\stackrel{P}{\longrightarrow}}$ $\newcommand{\rightarrowA}{\stackrel{a.s.}{\longrightarrow}}$

1. Infinitely often

Suppose that we have a sequence of subsets of some universal set. We shall define a new set from this sequence as follows.
Definition: Infinitely often (i.o.) Let $(A_n)$ be a sequence of subsets of some universal set. Then we define $$\{A_n\io\} = \cap_n \cup_{k\geq n} A_k.$$

EXAMPLE 1:  Suppose that $A_n = \left[ -\frac 1n, 1-\frac 1n \right].$ Find $\{A_n\io\}.$

SOLUTION: You can think of this sequence as $[-1,0]$ travelling towards $[0,1].$

Here $\cup_{k\geq n} A_k = \left[ -\frac 1n, 1\right).$

So $\{A_n\io\} = \cap_n \cup_{k\geq n} A_k = \cap_n [0, 1).$ ■

The definition may appear rather complicated. Let us digest itintuitively: If we take the intersection of all such unions of the form $\cup_{k\geq n} A_k,$ we get $\{A_n\io\}.$ Thus $\liminf A_n$ is the set all elements that are in all the $A_n$'s except for a finitely many, i.e., all the lements are are infinitely many of the $A_n\$'s. Hence the name infinite often.

The following result often proves handy when proving $P(A_n)\rightarrow 0.$

TheoremLet $(A_n)$ be a sequence of events with $P(A_n\io) = 0.$ Then we must have $P(A_n)\rightarrow0.$

A proof of this theorem is outlined in the exercises below.

1.1. Problem set

EXERCISE 1: If $A_n = [-n,n]$ find $\{A_n\io\}.$

EXERCISE 2: If $A_1 \subseteq A_2 \subseteq A_2\subseteq \cdots,$ then show that $\{A_n\io\} = \cup_n A_n.$

EXERCISE 3: If $A_1 \supseteq A_2 \supseteq A_2\supseteq \cdots,$ then is it true that $\{A_n\io\} = \cap_n A_n?$

EXERCISE 4: Let $B_n = \cup_{k\geq n} A_k.$ Show that $P(B_n)\rightarrow P\{A_n\io\}.$

EXERCISE 5: We can define a set called $\{A_n\ev\}$ (where $\ev$ is the abbreviation of eventually) as $$\{A_n\ev\} = \cup_n \cap_{k\geq n} A_k.$$ Do you see why we call it "eventually" (compare with the statement "We shall all be eventually dead.").

EXERCISE 6: Show that $\{A_n\ev\}\subseteq\{A_n\io\}.$

EXERCISE 7: Let $P\{A_n\ev\}=1.$ Show that $P(A_n)\rightarrow 1.$

EXERCISE 8:  We shall say $A_n\rightarrow A $ if $\{ A_n\ev\} = \{ A_n\io\} = A.$ Show that in this case $P(A_n)\rightarrow P(A).$ This is a generalisation of the theorem on continuity of probability that you have learned in Probability I.

EXERCISE 9: Let $(A_n)$ be a sequence of events with $P(\liminf A_n) = P(\limsup A_n) = p.$ Show that $P(A_n)\rightarrow p.$

2. Borel-Cantelli lemmas

First Borel-Cantelli lemma Let $(A_n)$ be a sequence of events in some probability space. If $\sum P(A_n) < \infty,$ then $P(A_n\io)=0.$

Proof: $$\begin{eqnarray*} P(A_n\io) & = & P(\cap_n\cup_{k\geq n} A_k)\\ & = & P( \lim_n\cup_{k\geq n} A_k)\\ & = & \lim_n P(\cup_{k\geq n} A_k). \end{eqnarray*}$$ Now $P(\cup_{k\geq n} A_k)\leq \sum_{k\geq n} P(A_k).$

Since $\sum P(A_n) < \infty$, hence $\lim_n \sum_{k\geq n} P(A_k) = 0.$

Hence the result. [QED]

Second Borel-Cantelli lemma Let $(A_n)$ be a sequence of independent events in some probability space. If $\sum P(A_n)= \infty,$ then $P(A_n\io)=1.$

Proof:Skipped.[QED]

The second lemma makes the assumption of independence. The following counterexample shows what may happen if this assumption were dropped.

EXAMPLE 2:  Let $U\sim Unif(0,1)$ and for $n\in{\mathbb N}$ let $$A_n = \left\{U\in\left(0,\frac 1n\right)\right\}.$$ Notice how we are defining all the $A_n$'s based on the same $U.$ Clearly, $A_n$'s are not independent. For instance, $P(A_n)=\frac 1n$, but $P(A_n|A_{n+1})=1.$

Here $\sum_n P(A_n) = \sum_n \frac 1n = \infty,$ but $P(A_n \io) = 0.$ ■

EXAMPLE 3: Show that if a monkey types a keyboard randomly, then it will type the exact text of Hamlet infinitely often.

SOLUTION: Let the exact text of Hamlet have $h$ letters. Consider packets of $h$ letters. Let $A_n$ be the event that the $n$-th packet is an exact copy of Hamlet. Then $A_n$'s are independent and $P(A_n)>0.$ So $\sum_n P(A_n) = \infty.$ Hence by the second Borel-Canteli lemma, $P(A_n\io)=1.$ ■

2.1. Problem set

EXERCISE 10: If a coin with $P(head)=0.01$ is tossed repeatedly (independently of course), then find the probability that there will be infinitely many heads.

EXERCISE 11: A fair coin is tossed independently infinitely many times. What is the chance that there would be infinitely many consecutive heads?

Hint:

Be careful: If $X_n$ denotes the outcome ofthe $n$-th toss, then $(X_1,X_2)$ is not independent of $(X_2,X_3)$.

EXERCISE 12: Let $(X_n)$ be any jointly distributed sequence of $Unif(0,1)$ random variables. Show that $P\left(X_n < \frac{1}{n^2} \io\right) = 0.$

EXERCISE 13: Let $(X_n)$ be an iid sequence of $Unif(0,1)$ random variables. Show that $P\left(X_n < \frac 1n \io\right) = 1.$

EXERCISE 14: Toss a fair coin repeatedly (independently). What is the probability that heads occur on all $n$-th tosses where $n$ is a square number?

Hint:

You really do not need the Borel Cantelli lemmas here.

EXERCISE 15:  Construct a sequence of events $(A_n)$ such that $\sum_n P(A_n) = \infty$ but $P(A_n\io) = 0.$

EXERCISE 16: Consider $[-1,1]$ as a 1-dimensional dart board, 0 being the bull's eye. A drunkard is throwing darts at this board hitting the board at any random points (all points equally likely). After the each attempt he gets a prize if he hits within $\frac 1n$ distance of the bull's eye (if it is the $n$-th attempt). What is the chance that he gets infinitely many prizes? [Assume infinite life for the guy, but don't drink to it!]

EXERCISE 17: Same problem as before except that $[-1,1]$ is replaced by the unit disc in ${\mathbb R}^2,$ the bull's eye being at the origin. What is the chance that he gets at leaast one prize? Wht is the chance he gets infinitely many prizes?

EXERCISE 18: However, if you want to just show that the monkey will surely write Hamlet at least once, then you really do not need to use the second Borel Cantelli lemma. You can prove it much more easily using continuity of probability. How?

Project: Almost all numbers are normal

Consider a number between 0 and 1. Consider its (unique) non-terminating decimal expansion. Pick any natural number $k$ and any of the $10^k$ digit patterns of length $k.$ Is any pattern more likely than another? For instance, in $\frac 13 = 0.33333...$ then pattern $333$ is the only pattern. If every patten in "equally frequent" in the number, then it is called a normal number. For instance, $\frac 13$ is not a normal number. Only a very few normal numbers are known. Nobody yet knows if $\pi$ or $e $ is a normal number. Yet, the surprising fact is that if you generate a random number from $Unif(0,1),$ then it is normal with probability 1. This project is about exploring this proof.

This project is of a more theoretical nature.
Theorem

Next we shall fix a single number $\omega\in(0,1)$ and check how many times any given digit (say 3) occurs in it.

Let $S_n(\omega) = $ number of times the digit 3 occurs among the first $n$ digits of $\omega.$ For instance of $\omega=\frac 13,$ then $S_n(\omega) = n.$

We are interested in the proportion of times a digit occurs in the expansion. Mathematically, we work with $\lim_{n\rightarrow \infty} \frac{S_n(\omega)}{n}$ (which may not exist). For instance, if $\omega=\frac 13,$ then the limit exists and equals $1.$ If $\omega=0.3,$ then the limit exists and equals $0.$

EXERCISE 19: Can you construct an expansion for which the limit does not exist?

Hint:

You can get one using only 0's and 3's.

Let $A_3 = $ set of all numbers for which the limit equals $\frac{1}{10}.$ What do you expect $P(A_3)$ to be? The following theorem provides the answer.
Theorem $P(A_n) = \frac{1}{10}.$

Proof: We shall prove it for $n=3.$ (the other cases are exactly similar).

[QED]

Then $\forall k\in\{0,1,2,...,9\}~~P(d_n=k\io) = 1.$ So, in particular, $P(0\mbox{ occurs }\io)=1.$ Hence, $P(0,1\mbox{ occur }\io)=1.$ Proceeding similarly, $P(\mbox{each digit occurs }\io)=1.$

Project: Random walk in 3D may not return A bird starts at the origin, and in each second moves one unit parallel to one of the three axes (the axis and direction chosen randomly). Is the bird sure to fly back to the origin sometime or other during the course of its random flight? The answer is "No!". But if we force the bird to stay in the $xy$-plane (i.e., vertical movements are not allowed), then the answer is "Yes". This project is about using the first Borel-Cantelli lemma to prove the general case.

It is a classical application of the Borel-Cantelli lemmas. Theoretical project. Not too difficult.