I got stuck in this question and I can't find a counterexample. I proved that this is true:
If ${|E|_e}<\infty\Rightarrow$ E is measurable $\Leftrightarrow{|E|_i}={|E|_e}$. Where ${|E|_i}$ and ${|E|_e}$ stand risp. for inner and outer measure.
Well, I cannot find a counterexample in the case when E has infinite outer measure. My book says it is not true but nothing comes to my mind. Everything I think of, has some closed subset with infinite measure so the equivalence still holds. To find one, I should find a set whose closed subsets have all finite measure or such that they are not measurable,right? Any ideas or suggestions?
P.s for those who do not know, the inner measure of E is defined by ${|E|_i}$= sup|F| where F$\subset$E is closed.
Best Answer
Here is a simple counter-exemple: a non-measurable set $E$ such that $|E|_i=|E|_e=+\infty$
Let $A$ be a non-measurable subset of $[0,1]$. Take $E=(-\infty,-1] \cup A$. Clearly $E$ is non-measurable. We also have that $$ |E|_i \geqslant |(-\infty,-1] |_i =+\infty$$ and $$ |E|_e \geqslant |(-\infty,-1] |_e =+\infty$$ So, we have $|E|_i=|E|_e=+\infty$.