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.
 A: 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.
A: 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.
