Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In preparation for the real analysis qualifying exam at my grad school, I've been working through the recommended textbook Modern Real Analysis, by Zimmer (it's free online here).

For the last few days, I've been trying to figure out problem 4.27 (on p124 of the text, which is p132 of the pdf), which asks us to prove the following:

Theorem. If $\{E_k\}_{k=1}^\infty$ is a sequence of Lebesgue measurable subsets of a compact set $K\subseteq \mathbb{R}^n$ such that $\inf_{k\geq 1} \lambda(E_k)>0$, then there's some point which belongs to infinitely many $E_k$'s (i.e. there's some point which belongs to $E_k$ for infinitely many $k$).

Definitions, Etc.

  • Lebesgue measure is denoted by $\lambda$.
  • The Lebesgue outer measure $\lambda^*$ is defined to be $$\lambda^*(A):= \inf\left\{\sum_{I\in\mathcal{S}} \mathrm{vol}(I) \right\},$$ where the infimum is taken over countable all covers $\mathcal{S}$ of $A$ by sets of the form $[a_1,b_1]\times \cdots \times [a_n,b_n]$ (which Zimmer calls closed $n$-dimensional intervals, or just closed intervals if the dimension is clear).
  • A set $A\subseteq\mathbb{R}^n$ is defined to be Lebesgue measurable if $$\lambda^*(A)=\lambda^*(A-S)+\lambda^*(A\cap S)$$ for any set $S\subseteq \mathbb{R}^n$.

My Attempt

If there are finitely many distinct $E_k$'s, this is trivial. So, assume there are infinitely many distinct $E_k$'s (I don't know if this assumption will be useful, but since we're working with compactness, I figured it might). From each $E_k$, choose a point $x_k$. Then because $K$ is compact, there's a subsequence $\{x_{k_j}\}_{j=1}^\infty$ of $\{x_k\}_{k=1}^\infty$ which converges to some point $x\in K$. At this point, I'm stuck. My gut tells me that this point $x$ (or some similarly constructed point) should be in infinitely many $E_k$'s, but I can't for the life of me figure out how to prove this.

Could someone point me in the right direction?

share|improve this question
1  
The gut is wrong; consider the characteristic functions and their sum. –  user31373 Jul 19 '12 at 3:16

2 Answers 2

up vote 1 down vote accepted

Suppose every point is contained in at most finitely many $E_k$, so the indicator functions $\chi_{E_k}:K\to \mathbb R$ converge to $0$ pointwise. Since $K$ has finite measure, we can apply Egorov's theorem, so for any $\epsilon>0$ we have a set $S\subseteq K$ such that $\lambda(K\setminus S)<\epsilon$ and $\chi_{E_k}|_S$ converges uniformly to $0$. But since $\chi_{E_k}|_S$ takes only the values $0$ and $1$, we must have that this sequence is eventually identically $0$ hence $E_k\cap S=\emptyset$ for sufficiently large $k$. But then $$\lambda(E_k)\leq \lambda(E_k\cap S)+\lambda(K\setminus S)=\epsilon$$ and since $\epsilon$ is arbitrary this violates the fact that $\inf_{k\geq 1} \lambda(E_k)>0$. Thus some point is contained in infinitely many $E_k$.

share|improve this answer
    
Or simply dominate by 1. –  user31373 Jul 19 '12 at 3:44
    
@LeonidKovalev Yes, I suppose convergence in the $L^2$ norm is good enough here. I just usually go with Egorov as soon as I see I can because it's a strong technique. –  Alex Becker Jul 19 '12 at 3:51
1  
I did not mean $L^2$. Dominated convergence says the integrals converge to 0, contrary to the assumption with $\lambda$. Yet another version: Fatou applied to $1-\chi_{E_n}$. –  user31373 Jul 19 '12 at 4:03
    
Isn't the convergence of the integrals to $0$ in this case equivalent to convergence of the functions to $0$ in the $L^2$ norm? –  Alex Becker Jul 19 '12 at 4:15
    
It is. Also equivalent to convergence in the $L^{\pi}$ norm. It's just that nothing in this problem makes me think of either. –  user31373 Jul 19 '12 at 4:39

We can use the following: Given inf E_k is larger. Make use of limsup of the sets. Let A_1 be union from 1 to infinity of E_k's this is of course geq E_1 hence infimum > 0 Let A_2 be union from 2 to infinity of E_k's this is of course geq E_2 hence infimum > 0 and so on

Take the intersection of those. Those are the set of points that are to be found in infinitely many sets E_k. This intersection being larger than or equal to infimum has positive measure meaning it contains uncountably many points.

share|improve this answer

Your Answer

 
discard

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.