Measurable Sets. Need help?

  • #1
sutupidmath
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Problem. Let E be the closed unit square. Prove that every open subset of E is measurable.


I know that one way to show that a set, say A, is measurable is to show that its outer and inner measure coincide; another way is to exibit an elementary set B such that

[tex] \mu(A\Delta B)< \epsilon.[/tex]

However, I am not sure where to start. Any hints would be appreciated?
 

Answers and Replies

  • #2
mathwonk
Science Advisor
Homework Helper
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i don't know, maybe use cubes and compactness? what is your definition of measurable?
 
  • #3
sutupidmath
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i don't know, maybe use cubes and compactness? what is your definition of measurable?

A set is defined to be measurable if its outer measure coincides with the inner measure. If A is a subset of the unit square E, we say that

[tex] \mu^*(A)=inf\{\sum_{k}m(P_k): A\subset \bigcup_{k}P_k, \mbox{ it is taken over all subcovers and } P_k \mbox{ are rectangles} \}[/tex],

is the outer measure of A.

On the other hand the inner measure is defined as

[tex]\mu_*(A)=1-\mu^*(E-A)[/tex]

So, a set A is measurable if

[tex] \mu^*(A)=\mu_*(A).[/tex]

But then there is a theorem that says that a set A is measurable iff there exists some elementary set B, such that given any epsilon>0, we have

[tex]\mu(A\Delta B)<\epsilon. [/tex]

Also, an elementary set B, is defined to be a finite collection of pairwise disjoint rectangles.
 
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  • #4
sutupidmath
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Anyone?
 
  • #5
disregardthat
Science Advisor
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How about proving it for arbitrary epsilon balls? As QxQ intersected with any open subset of the unit square is dense, you can provide a countable cover of epsilon balls of any open subset of E. Countable unions of measurable sets are measurable.
 

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