Excision property false when set is not measurable. From Royden &amp; 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.
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 
$$m^\*(A)&lt; m^\*(A\cup E)+ m^\*(A\cup E^c)$$
In specific for a open set $O$ covering $E$, with $O\cup E=E$, we must have 
$m^\*(O)&lt; m^\*(O\cup E)+ m^\*(O\cup E^c)     .....$
$m^\*(O)&lt; m^\*(E)+ m^\*(O~E) ....$
$m^\*(O)-m^\*(E) &lt;  m^\*(O\sim E)$
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 
$$m^\*(A)&lt; m^\*(A\cup E)+ m^\*(A\cup E^c)$$
In specific for a open set $O$ covering $E$, with $O\cup E=E$, we must have 
$m^\*(O)&lt; m^\*(O\cup E)+ m^\*(O\cup E^c)     .....$
$m^\*(O)&lt; m^\*(E)+ m^\*(O~E) ....$
$m^\*(O)-m^\*(E) &lt;  m^\*(O\sim E)$