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Consider the probability model $(\Omega, \mathcal{F}, P)$ where $\Omega = [0,1]$, $\mathcal{F}$ is the Borel $\sigma$-algebra on $[0,1]$ and $P$ is the uniform measure on $[0,1]$.

Let $E_1, E_2, \dots$ be a sequence of independent events in $\mathcal{F}$ such that $P(E_n) = 1/n$.

Since $\sum P(E_n) = \infty$, by the second Borel–Cantelli lemma (Wikipedia), $$ P(\bigcap_{k=1}^\infty \bigcup_{n=k}^\infty E_n) = 1. $$

My question is:

Is there any explicit example of such $E_n \subset [0,1]$?

The most famous example of independent events on $[0,1]$ is the dyadic intervals: Define

$D_1 = [0, 2^{-1}]$

$D_n = 2^{-1}D_{n-1} + (2^{-1}+2^{-1}D_{n-1}) $

$(D_n)_{n =1}^\infty$ is a sequence of independent events. However, for this example, $P(D_n) = 1/2^n$.

Is there any explicit example for $P(E_n) = 1/n$ ?

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  • $\begingroup$ Just take an infinite sequence $X_n$ of independent random draws from $\Omega$ and let $E_n$ be the event $X_n\in [0,1/n]$. $\endgroup$
    – Alex B.
    Oct 31, 2016 at 11:29
  • $\begingroup$ @AlexB.That needs an explicit measure-preserving map from $[0,1]$ to $[0,1]^\infty$, to answer the question as it is... Otherwise, take the canonical map from $[0,1]$ to $\{0,1\}^\infty$ given by dyadic expansion, but then it would be hard to match $P(E_n)=1/n$ exactly (while having it approximately is easy). $\endgroup$ Oct 31, 2016 at 12:26

4 Answers 4

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All irrational numbers in $(0,1)$ have unique representation as $\sum_{k=2}^\infty c_k/k!$, where $c_k\in \{0,1,\dots,k-1\}$. 'Digits' $c_k$ are independent, so you may choose the events $E_n$ as '$c_n=0$', for example.

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  • $\begingroup$ That's a nice one $\endgroup$
    – R W
    Oct 31, 2016 at 17:40
  • $\begingroup$ Simple and elegant $\endgroup$
    – Sam Wong
    Nov 22, 2020 at 7:44
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Construction of such an example essentially amounts to an explicit decomposition of the unit interval $(\Omega,P)$ into a product of probability spaces. For instance, one can take an isomorphism between $(\Omega,P)$ and its infinite countable product by itself $(\Omega,P)\times(\Omega, P)\times \dots$. [Since $(\Omega,P)$ is a countable product of the dyadic space $\{0,1\}$ endowed with the uniform measure by itself, an explicit rearrangement of binary digits will then provide a required isomorphism.] Now, as it was suggested in Alex.B's comment, you just take in each of the multipliers a set of any measure you like.

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It is awkward to construct an event of prob 1/17, e.g. out of dyadic, or even decimal digits. As you want a borel cantelli example in the unit interval, why don't you change your probabilities instead to make the point the same but the construction easier. Make e.g. $E_{17}$ a set of probability $[1000/17]/1000 = .058$. Then your event might be decimal digits 49 -51 when placed in that order are a number < 58.

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  • $\begingroup$ It is not so awkward. Say, consider the minimal $k$ for which not all the digits from $(8k+1)$-st to $(8k+8)$-th are zeroes. The probability that the dyadic number formed by these 8 digits is divisible by 17 equals $1/17$. $\endgroup$ Oct 31, 2016 at 18:32
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Let $(X_i)_{i\geq 1}$ be any sequence of independent, identically distributed random variables such that $P(X_i = x)=0$ for all $x$. Such sequence can be built on the universe $[0,1]$ itself because any Borel standard space with a non-atomic probability measure is isomorphic to $[0,1]$ together with the Lebesgue measure. Define $$ E_n = \{\omega \mid \forall \, k <n, \, X_k(\omega) < X_n(\omega)\}. $$ If the event $E_n$ occurs, we say that there is a record at time $n$. A standard exercice in probability theory asks to show that the $E_n$ are independent and that $P(E_n) = 1/n$. By Borel-Cantelli, this implies that there are infinitely many records almost surely.

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