Proving E is Measurable with Compact Sets

[jdcasey9]

Homework Statement

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

Homework Equations

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

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\bigcupE₂, E₁\bigcapE₂ 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\bigcap K is measurable by Lemma 16.14 (Carothers).

Assume E\bigcapK is measurable for compact K. Since K is compact, it is necessarily closed. So E\bigcapK is also compact and closed because a closed subset of a compact set is compact. Then, because E\bigcapK 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) < \epsilon.

For all integers n, F_n (closed set), G_n )(open set) with F_n\subset E\bigcapK\subsetG_n such that m*(G_n\F_n) < 2^-n\epsilon where G = \bigcup G_n is an open set containing E and E=F=\bigcupF_n is a closed set. So, G\F \subset \bigcup (G_n\F_n) and m*(G\F) \leq \Sigma m*(G_n\F_n) < \Sigma 2^-n\epsilon = 3\epsilon. So E is measurable.

Is this a sound argument?

 

[micromass]

Hello jdcasey09!

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

Assume E\bigcapK is measurable for compact K. Since K is compact, it is necessarily closed.

This is not true. E\cap K 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 E\cap K 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...