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.

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    $\begingroup$ 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? $\endgroup$ Jul 17, 2022 at 22:00

2 Answers 2


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 \begin{equation} \sum_{i=1}^{\infty} P(A_n) = \infty, \end{equation} then \begin{equation} P \left( \bigcap_{i=1}^{\infty} \bigcup_{n=i}^{\infty} A_i \right) = 1 \end{equation}

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, \begin{equation} \sum_{i=1}^{\infty} P(A_i)^2 < \infty \end{equation} and, by our Theorem, \begin{equation} P \left( \bigcap_{i=1}^{\infty} \bigcup_{n=i}^{\infty} A_i \right) = 1 > 0. \end{equation}

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.


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