Proving E is Measurable with Compact Sets

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jdcasey9
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Homework Statement



Prove that E is measurable if and only if E [tex]\bigcap[/tex] K is measurable for every compact set K.

Homework Equations


E is measurable if for each [tex]\epsilon[/tex] < 0 we can find a closed set F and an open set G with F [tex]\subset[/tex] E [tex]\subset[/tex] G such that m*(G\F) < [tex]\epsilon[/tex].

Corollary 16.17: Open sets, and hence also closed sets, are measurable.

Lemma 16.14: If E1 and E2 are measureable sets, then so are E1[tex]\bigcup[/tex]E2, E1[tex]\bigcap[/tex]E2 and E1\E2.


The Attempt at a Solution



Assume E is measurable. Since a compact set is always closed, K is closed and is measurable by Corollary 16.17 (Carothers). Then, because E and K are both measurable, E[tex]\bigcap[/tex] K is measurable by Lemma 16.14 (Carothers).

Assume E[tex]\bigcap[/tex]K is measurable for compact K. Since K is compact, it is necessarily closed. So E[tex]\bigcap[/tex]K is also compact and closed because a closed subset of a compact set is compact. Then, because E[tex]\bigcap[/tex]K is measurable, it is between a closed set F and an open set G. Since it is closed, it must be equal to F for m*(G\F) < [tex]\epsilon[/tex].

For all integers n, Fn (closed set), Gn )(open set) with Fn[tex]\subset[/tex] E[tex]\bigcap[/tex]K[tex]\subset[/tex]Gn such that m*(Gn\Fn) < 2-n[tex]\epsilon[/tex] where G = [tex]\bigcup[/tex] Gn is an open set containing E and E=F=[tex]\bigcup[/tex]Fn is a closed set. So, G\F [tex]\subset[/tex] [tex]\bigcup[/tex] (Gn\Fn) and m*(G\F) [tex]\leq[/tex] [tex]\Sigma[/tex] m*(Gn\Fn) < [tex]\Sigma[/tex] 2-n[tex]\epsilon[/tex] = 3[tex]\epsilon[/tex]. So E is measurable.

Is this a sound argument?
 
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Hello jdcasey09! :smile:

I'm afraid there is a flaw in the argument:

jdcasey9 said:
Assume E[tex]\bigcap[/tex]K is measurable for compact K. Since K is compact, it is necessarily closed.

This is not true. [tex]E\cap K[/tex] is simply a measurable subset of K, it can be everything. thus it is not necessarily closed!

For example, if K=[-2,2] and E=]-1,1[, then [tex]E\cap K[/tex] is a measurable subset of K, but it isn't closed.

I also see that you don't specify what K is in your proof. That is, you keep K general. This is good, but eventually you'll need to give a specific form for K...