Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 1 down vote accepted

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 $k\geq 1$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.