# Let $A_i$ be measurable and $\sum_{i = 1}^\infty \mu (A_i)^2 < \infty$. Is $\mu( \bigcap_{ i = 1}^\infty \bigcup_{m = i}^\infty A_m) = 0$?

Let $$\mu$$ be an arbitrary measure on a measure space $$X$$, so that we have the following finite sum $$\sum_{i = 1}^\infty \mu (A_i)^2 < \infty.$$ I am asked to prove or disprove the following: $$\mu( \bigcap_{ i = 1}^\infty \bigcup_{m = i}^\infty A_m) = 0.$$

I am not sure how to go about this. I think it may not be true but all help in either direction is greatly appreciated.

Since the measures are squared, I thought to take $$\mu(A_i) = 1/i$$ and exploit the fact that the harmonic sum diverges somehow.

On the other hand, we don't know that one of the measures of $$\bigcup_{m = i}^\infty A_m$$ is finite, so we can't apply some continuity property to exploit the fact that $$\lim_{i \to \infty} \mu(A_i) = 0$$.

Thanks for any help or hints.

• I like your idea. Now, can you use it to come up with a sequence of intervals that covers the real line infinitely many times? Jul 17, 2022 at 22:00

You should take a look at the Borel-Cantelli lemma, from Probability Theory, specially the so called Second Borel-Cantelli lemma. We state it here:

$$\textbf{Theorem}$$ Let $$(\Omega, \mathcal{F}, P)$$ be a probability space. If the sequence of $$\it{independent}$$ events $$(A_n)_{n\in\mathbb N}$$ satisfies $$$$\sum_{i=1}^{\infty} P(A_n) = \infty,$$$$ then $$$$P \left( \bigcap_{i=1}^{\infty} \bigcup_{n=i}^{\infty} A_i \right) = 1$$$$

With this at our disposal, we can disprove your proposition by, for example, constructing a probability space and a sequence of independent events with $$P(A_n) = 1/n$$ for every $$n \in \mathbb{N}$$, as your intuition told you.

To do this formally would be very daunting, but the following simple experiment exemplifies it is possible:

The experiment consists of, for each $$n \in \mathbb{N}$$, taking a $$n$$-sided fair die and throwing it.

Now, let $$A_n$$ be the event "The $$n$$th throw resulted in a 1". Obviously those events are independent, and they satisfy $$P(A_n) = 1/n$$ for every $$n$$. Hence, $$$$\sum_{i=1}^{\infty} P(A_i)^2 < \infty$$$$ and, by our Theorem, $$$$P \left( \bigcap_{i=1}^{\infty} \bigcup_{n=i}^{\infty} A_i \right) = 1 > 0.$$$$

I don't know how familiared you are with the measure-theoretic formulation of Probability so, if you have any questions, let me know.

Take $$X$$ to be $$\mathbb R$$ with the usual Lebesgue measure. We construct $$A_n$$ such that $$\mu(A_n)=\frac{1}{n}$$. Basically, we try to cover $$[0, k]$$ for each $$k\in\mathbb N$$, and then start all over again to cover $$[0, k+1]$$, etc. Formally, for each $$k\in\mathbb N$$, we may inductively introduce $$n_k$$ such that $$\begin{cases} n_1=1 \\ n_k=\min_m\{\sum_{i=n_{k-1}}^m\frac{1}{n}\ge k\}\end{cases}$$

Note that the divergence of the harmonic series is needed for $$n_k$$ to be defined.

Now for each $$n$$, we define $$A_n=[a_n, a_n + \frac{1}{n}]$$, where $$a_n=0$$ if $$n=n_k$$ for some $$k$$, and $$a_n = a_{n-1} + \frac{1}{n-1}$$ otherwise.

It should be clear that each of $$[0, \infty)$$ appears in the family $$\{A_n\}$$ infinitely often, but the interval has inifnite measure.