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.

Let $X \subseteq \mathbb{R}$. A subset $E \subseteq \mathbb{R}$ is called a hull of $X$ if

  1. $E$ is measurable

  2. $X \subseteq E$

  3. If $F$ is any measurable set such that $X \subseteq F$, then $E$\ $F$ is a zero set

A hull of $X$ should be thought of as a sort of "smallest" measurable set containing $X$. This is made precise in the problem.

(a) Prove that every subset of $\mathbb{R}$ has a hull

(b) Prove that if $E_{1}$ and $E_{2}$ are both hulls of $X$, then $E_{1}$\ $E_{2}$ and $E_{2}$\ $E_{1}$ are zero sets. Thus the hull of $X$ is unique up to zero sets

(c) Let E be a hull of X. Prove that $m(E)=m^{*}(X)$

The hint says to consider the case $m^{*}(X)< \infty$, then consider the case let $R_i=(i,i+1)$ for $i \in \mathbb{Z}$. Let $X_i=X \cap R_i$. Then $X=\cup_{i \in \mathbb{Z}} X_{i} \cup \mathbb{Z}$. I need help to prove this.

share|improve this question
    
Haven't you posted this question before? –  Frank Apr 1 at 13:36
    
Yes, but I changed the question this time. I hope someone can help me understand the problem. –  user220055 Apr 1 at 13:38
    
For part (a) can't you just take the intersection of all measurable sets containing $X$? –  Frank Apr 1 at 13:39
    
Ok, I guess you got this. Since you said it's obvious. –  Frank Apr 1 at 13:40
    
no, I kind got b, not a. –  user220055 Apr 1 at 13:42
show 10 more comments

2 Answers 2

up vote 1 down vote accepted

Let $X\subset \mathbb{R}$ with finite measure, then there exist open sets $A_n\supset X$ such that $$m^*(A_n) - m^*(X) \leq \frac{1}{n}.$$

Define $A = \cap A_n$, then $A$ is measurable and $A\supset X$. To show (c) $$m^*(A) - m^*(X) \leq m^*(A_n) - m^*(X) \leq \frac{1}{n}$$ for each $n$, we have that $m^*(A) = m^* (X)$.

Let $F\supset X$ be measurable, then $$m^*(A) = m^*(A\cap F) + m^*(A\cap F^c) \geq m^*(X) + m^*(A\cap F^c)$$ which gives $$0 \geq m^*(A\cap F^c).$$

For part (b), use the same argument as above, take $F$ to be another Hull of $X$, all the inequality would still work.

For $X\subset\mathbb{R}$ with $m^*(X) = \infty$, define $X_i = X\cap([-i, -i+1)\cup [i-1, i))$, then $X = \cup_{i=1}^\infty X_i.$ By previous arguement, there exists $A_i$ for each $X_i$, with all $A_i$ disjoint. Define $A = \cup_i A_i$, we see that $$m^*(A) = \sum_i m^*(A_i) = \sum_i m^*(X_i) \geq m^*(X) = \infty.$$

share|improve this answer
    
There are two cases in part (c). Is (c) based on (a) and (b)? –  user220055 Apr 1 at 15:44
    
(c) is not really based on (a) and (b). You can say (c) is a property that holds when the set $A$ is a hull of $X$. –  Xiao Apr 1 at 15:48
    
ok, thank you.. –  user220055 Apr 1 at 15:49
    
Oh yes, I forgot, the case I showed is for $\mu(X) < \infty$, can you figure out the rest using the hint in (c)? –  Xiao Apr 1 at 15:52
    
I probably can't figure it out...can you show that part please? –  user220055 Apr 1 at 15:55
show 4 more comments

a): Hulls occur very often in differents fields of mathematics. Usually, you have some set $M$ which doesn't fullfill a particular property (measureable, in this case), and you're looking for the *smallest (in some sense) set which includes $M$ and does have the desired property. For example, in a vector space, a natural definition for the hull of a set of vectors $M$ is the smallest subspace with contains $M$. Now, if the propertery in question (being a subspace, in the case of vector spaces) is preserved under intersections, then this is easy - just define $$ \textrm{hull } M := \bigcap_{X \supset M, \textrm{$X$ has desired property}} X \text{.} $$

The problem in your case is that it's not true that the intersection of arbitrary measurable sets is measureable. But the intersection of arbitrarily many closed sets is closed, and every closed set is measurable. So a natural idea for finding the measurable hull of an arbitrary set $X \subset \mathbb{R}$ is to set $$ \textrm{hull } X := \bigcap_{X \subset T \subset \mathbb{R}, \textrm{$T$ closed}} T \text{.} $$ It's clear that $\textrm{hull } X$ is closed, and hence measurable. What remains to show is that if $F \supset X$ is measurable, $\mu(\textrm{hull } X \setminus F) = 0$. For that, you'll need to use that the closed sets in some way (which, exactly?) generate all the measurable sets.

c): If you can show that for the special hull constructed in (a) that $m^*(\textrm{hull } X) = m^*(X)$, then the same must hold for any hull $E$ - just invoke (b).

share|improve this answer
1  
Intersection of arbitrarily many open sets does not have to be open :P –  Xiao Apr 1 at 15:14
    
@Xiao Ups! Should be closed, of course! –  fgp Apr 1 at 15:16
    
I cannot figure out how to prove μ(hull X∖F)=0. any hint? –  user220055 Apr 1 at 15:30
    
@user220055 You need to approximate $F$ with a sequence of closed sets... –  fgp Apr 1 at 15:34
    
I need more explanation... –  user220055 Apr 1 at 15:41
show 2 more comments

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.