Proving Finite Outer Measure Inequality

[Artusartos]

Homework Statement

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 measure and for which

m*(O~E) > m*(O) - m*(E)

Homework Equations

The Attempt at a Solution

This is what I did...

$m^*(O) = m^*((O \cap E^c) \cup m^*((O \cap E)) \leq m^*(O \cap E^c) + m^*(E)$

So...

$m^*(O) - m^*(E) \leq m^*(O \cap E^c)$

But I'm confused about the equality, because the one in the question is a strict inequality

 

[Fredrik]

Homework Statement

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 measure and for which

m*(O~E) > m*(O) - m*(E)

Homework Equations

The Attempt at a Solution

This is what I did...

$m^*(O) = m^*((O \cap E^c) \cup m^*((O \cap E)) \leq m^*(O \cap E^c) + m^*(E)$

So...

$m^*(O) - m^*(E) \leq m^*(O \cap E^c)$

But I'm confused about the equality, because the one in the question is a strict inequality

What you have done is just to prove that if there's an open O that contains E, then the last inequality in your post is satisfied. You have done nothing to prove that such a set exists. Of course, the entire space is always an open set, so that's not really an issue, at least not when the space has finite measure. But this still suggests that you need to find a special kind of open set, not an arbitrary one. And you should expect to have to use the definition of "measurable" to find it.

It would have been a good idea to include the definition of "measurable" under "relevant equations".