# An approach to Borel-Cantelli for the $l^p$

Let $\mu$ be a non-negative measure and $\{E_{k}\}$ a sequence such that $\sum \mu(E_k)^p<\infty$ then show that $F=\lim \ \sup E_{k}=\cap_{k=1}^{\infty}\cup_{n\geq k}E_n$ has $\mu$ measure zero.

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Do you require $p\in (0, 1]$, or any $p$? –  William Feb 16 '12 at 3:57
The interesting case will appear when $p>1$. That case is justified by Borel-Cantelli. –  checkmath Feb 16 '12 at 4:03
Exactly, that's why I was making sure. –  William Feb 16 '12 at 4:10
Well, I am thinking about a counter example for p=2, which may be adequate to the others. The example is take the series $\sum \frac{1}{n}$ and then make $E_1=[0,1]$ cover itself, $E_{2}=[0,1/2]$, $E_{3}=[1/2,5/6]$, $E_{4}=[5/6,13/12]$, $E_{5}=[0,1/5]$... with $\mu(E_{k})=1/k$ and note that $[0,1]\subset F$. –  checkmath Feb 16 '12 at 5:12
There is a counter example... I'm typing it up right now. –  William Feb 16 '12 at 5:17

When $p\in (0, 1]$, this obviously reduces to the classical Borel-Cantelli lemma. When $p > 1$, the result is not true in general. For a counterexample, consider the following construction (we take $\mu$ to be the Lebesgue measure on $\mathbb{R}$).

So basically the idea is to cover $(0,1)$ by a sequence $\{F_j\}$, such that $\mu(F_j) = 1/j$, while for arbitrarily large $k$, $\{F_j\}_{j\geq k}$ still cover $(0,1)$.

For any $n\geq 1$, let $\mathcal{C}_n = \{E_{i,n}\}_{i\geq 1}$ be a finite open cover of $(0,1)$ such that $\mu(E_{i,n}) = 1/{(n-1+i)}$ (which can of course be constructed since $\sum_{k=1}^\infty 1/k$ diverges). Now construct a countable sequence of sets recursively as follows. First take $G_1 = C_1$. For $j > 1$, if $m$ is such that $\min_{E\in C_{j-1}}\mu(E) = 1/m$, set $G_j = C_{m-1}$. Let $F = \bigcup_{j}G_j$. Now we have:

$$\sum_{E\in F}\mu(E)^2 < \infty.$$

Enumerate elements in $F$ as $F_1, F_2, \dots$ in order of decreasing measure. Then we get

$$(0,1)\subset \bigcap_{j\geq k}F_j,$$

for every $k\geq 1$.

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PS. This works for any $p > 1$, since instead of $\sum_{E\in F}\mu(E)^2$, we now have, in general, a convergent $p$-series $\sum_{E\in F}\mu(E)^p$. –  William Feb 16 '12 at 5:39