Measurable Sets: Proving Open Subsets of Closed Unit Square are Measurable

[sutupidmath]

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

\mu(A\Delta B)< \epsilon.

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

 

[mathwonk]

i don't know, maybe use cubes and compactness? what is your definition of measurable?

 

[sutupidmath]

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

\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} \},

is the outer measure of A.

On the other hand the inner measure is defined as

\mu_*(A)=1-\mu^*(E-A)

So, a set A is measurable if

\mu^*(A)=\mu_*(A).

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

\mu(A\Delta B)<\epsilon.

Also, an elementary set B, is defined to be a finite collection of pairwise disjoint rectangles.

 

[sutupidmath]

Anyone?

 

[disregardthat]

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.