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Let $(X,M,\mu)$ be a measure space and let $E_{n}\in M$. Under the assumption that $\sum\mu(E_{n})<\infty$, we can get $\mu(\limsup E_{n})=0$. Now I am trying to prove or disprove that the conclusion follows if the assumption is replaced by $\sum\mu(E_{n})^{2}<\infty$.

In my attempt to disprove it, I wanted $\sum\mu(E_{n})^{2}<\infty$ but $\sum\mu(E_{n})=\infty$ so I tried with intervals of length $\frac{1}{n}$ (either taking $E_{n}=(0,\frac{1}{n})$ or the corresponding closed interval, or picking the $E_{n}$'s to be disjoint) but I end up with the $\limsup$ being either empty or a singleton. On the other hand, I don't know what I can use to prove it.

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up vote 3 down vote accepted

I'll try to construct a counterexample using the ideas you have, and then we can see if someone can come up with a cleaner one.

Take $X=[0,1]$ with Lebesgue measure. Let $S_n$ denote the $n$th partial sum of the harmonic series. Define $E_n$ as follows:

(i) $E_1 = [0,1]$.

(ii) For $n>1$, $E_n = [S_{n-1}, S_n]$, where the endpoints are interpreted modulo 1. If $S_{n-1}>S_n$, set $E_n = [0, S_n] \cup [S_{n-1}, 1]$. Basically, we're taking intervals of length $\frac{1}{n}$ and wrapping them around $[0,1]$ over and over, starting at the endpoint of the previous $E_n$.

By the divergence of the harmonic series, every point in $[0,1]$ is represented infinitely many times throughout the $E_n$, so $\limsup E_n = [0,1]$. But $\sum \mu(E_n)^2 < \infty$.

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That's an interesting example and it makes sense. But as you said, let's see if someone else has a different one. – user44532 Dec 28 '12 at 16:21

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