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From Royden & Fitzpatrick's Real Analysis, Problem 19 section 2.4:

"Let E have finite outer measure. Show that if E is not measurable, then there is an open set O containing E that has finite outer measure and for which m*(O/E) > m*(O) - m*(E)."

My process goes like this:

1) m(E) is finite => E is bounded => There exists open set O containing E such that for all e1>0, m(O)-m(E)

2) E not measurable => there exists some e2>0 such that there exists an open set Q containing E and m(Q/E)>e2.

If I could somehow show that for the same e>0, Q=O then I would have the property immediately. But I cannot see any way that I can justify equating O with Q, and I do not see any other angle of approaching the problem, anyone can offer any hints or answers, I'd much appreciate it.

Mike
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1 Answers1

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The contrary to say $E$ is measurable if for a set $A$ $$m^\*(A)\geq m^\*(A\cup E)+ m^\*(A\cup E^c)$$ is that for every set we have $$m^\*(A)< m^\*(A\cup E)+ m^\*(A\cup E^c)$$ In specific for a open set $O$ covering $E$, with $O\cup E=E$, we must have $m^\*(O)< m^\*(O\cup E)+ m^\*(O\cup E^c) .....$ $m^\*(O)< m^\*(E)+ m^\*(O~E) ....$ $m^\*(O)-m^\*(E) < m^\*(O\sim E)$

Nosrati
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José De La Torre
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    Please, improve the formatting of your answer. Like in the answer, you forgot to use "$" for math equations. – Filburt Mar 11 '17 at 23:39