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showing $mu(limsup_{ntoinfty} E_n) = 0.$

Mathematics Asked by shiloh.otis on January 21, 2021

Let $(X,mathcal{M},mu)$ be a measure space. Suppose $E_nin mathcal{M}$ such that $$sum_{n=1}^infty mu(E_n) < infty$$ show $mu(limsup_{ntoinfty} E_n) = 0.$

Also, how can I prove or give a counterexample of the conclusion if the hypothesis is replaced with $$sum_{n=1}^infty mu(E_n)^2 < infty$$

Can I prove this by Fatou’s Lemma? Thank you

3 Answers

As Bernard comments, the first part is actually the statement of Borel-Cantelli lemma.

For the second part, as Stella mentions, we may want to find a series which is not convergent but the series consists of square of its terms is convergent. The series $sum_{n=1}^infty frac{1}{n}$ will satisfy this condition.

So we need to construct a sequence of sets ${E_n}$ s.t. $mu(E_n)=frac{1}{n}, forall nge1.$ Also, we want that $mu(limsup_{ntoinfty} E_n) neq 0.$

To construct a concrete example, we may need a proposition.

Proposition: For every real number x, there exists a sequence $x_1,x_2,x_3,...$ of integers such that $$x=x_1+frac{x_2}{2!}+frac{x_3}{3!}+cdotcdotcdot ,,$$ where $x_1$ can be any integer, but for $nge2,x_nin{0,1,...,n-1}.$ Furthermore, if we require that the partial sums be strictly smaller than x, then such a representation is unique.

A proof can be found here: https://blogs.ams.org/mathgradblog/2017/09/20/real-numbers-base-factorials-by-product/

From the proposition and its proof, it's easy to see that, if $xin(0,1)$, being an irrational number, then we can find an unique sequence $x_1,x_2,x_3,...$ s.t. $x_1=0$ and $$x=frac{x_2}{2!}+frac{x_3}{3!}+cdotcdotcdot ,,$$ with $x_nin{0,1,...,n-1},forall nge2.$

Also note that, if $x$ is an irrational number, then the corresponding series will consist of infinitely many terms. (N.B. This is crucial to our construction, which makes sure the construction process will not terminate in finite steps.)

Our example will be given in the language of probability space, but in essence, it can be easily converted into the language of general measure theory.

We define the event $E_1={text{pick a random irrational number $x$ in $(0,1)$, $x_1$ is $0$}. }$. Then it's clear that $P(E_1)=1.$

For each $nge 2,$ define event $E_n={text{pick a random irrational number $x$ in $(0,1)$, $x_n$ is $0$}. }$. Then it's also clear that $P(E_n)=frac{1}{n}.$

Note that these events are mutually independent.

By the second Borel-Cantelli lemma(c.f. https://en.wikipedia.org/wiki/Borel–Cantelli_lemma#Converse_result), we conclude that $P(limsup_{ntoinfty} E_n)=1neq 0.$

So, even though $sum_{n=1}^infty mu(E_n)^2 < infty$, we may still have $mu(limsup_{ntoinfty} E_n)=1neq 0.$

$$tag*{$blacksquare$}$$

Correct answer by Sam Wong on January 21, 2021

Proof of the first part:

The result you mention is the first Borel-Cantelli Lemma. First recall that $$limsup_n E_n = bigcap_{n=1}^infty bigcup_{k=n}^infty E_k.$$ By continuity from above (check the conditions as an exercise) and sub-additivity, we have $$muleft(limsup_n E_nright) = lim_nmuleft(bigcup_{k=n}^infty E_kright)leq lim_n left(sum_{k=n}^infty mu(E_k)right).$$ Note that last term is the tail of a convergent series, hence it converges to zero.

For the second part see Sam Wong's solution.

Answered by varpi on January 21, 2021

It is a very general and very useful theorem of calculus that if $sum a_i < infty$ then it must be that $limsup a_i = 0$. To get your result from this theorem, simply apply subadditivity.

For the counter example, can you think of a series such that $sum a_i^2$ converges and $sum a_i$ diverges?

Answered by Stella Biderman on January 21, 2021

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