Non-measurability of a set (Outer Measure) Suppose that E ⊂ [0,1] and E is not measurable.  Show that 
m*(E) + m*([0,1] - E) > 1.  
I found a similar relation here; however, I am having difficulty understanding how to show this relationship.  Any guidance would be helpful.
http://math.bard.edu/belk/math461/NonMeasurableSets.pdf
 A: I suppose $E$ is Vitalli. Let show that $m^*([0,1]\backslash E)=1$. Since $E$ is non-measurable, $m^*(E)>0$, and thus, we'll have $$m^*(E\backslash [0,1])+m^*(E)>1.$$ 
Suppose by contradiction that $m^*(E)<1$. Then, there is an open set $\mathcal O\supset E$ s.t. $m(\mathcal O)<1$. In particular, $A:=[0,1]\backslash \mathcal O\subset E$ and $m(A)>0$. We know that $\{E+q\}_{q\in \mathbb Q\cap [-1,1]}$ are disjoints (see lemma 3 of your link). Therefore,
$$3=m([-1,2])\geq m\left(\bigcup_{q\in \mathbb Q\cap [-1,1]}E+q\right)=\sum_{q\in \mathbb Q\cap[-1,1]}m(E+q)=\sum_{q\in\mathbb Q\cap [-1,1]}m(E)=\infty ,$$
which is a contradiction. Therefore $m^*([0,1]\backslash E)=1.$
A: The outer measure of a set is defined as:
$$
m^*(E) = \inf\Bigg\{ \sum_{i=1}^n l(I_i) : E \subset \bigcup_{i=1}^n I_i \Bigg\}.
$$
Now, it is clear that $m^*(E) + m^*([0,1] \backslash E) \geq m^*([0,1])$, because if the intervals $I_i$ cover $E$ and the intervals $J_i$ cover $[0,1] \backslash E$, then $\{I_i\} \cup \{J_i\}$ cover $[0,1]$, hence every cover of $E$ and $[0,1] \backslash E$ also covers $[0,1]$, so $m^*(E) + m^*([0,1] \backslash E) \geq m^*([0,1])$.
Now, if $E$ were measurable, then by the definition of being measurable, $m(E) + m([0,1] \backslash E) = m([0,1])$. Since $E$ is not measurable, it follows that $$m^*(E) + m^*([0,1] \backslash E) > m^*([0,1]) = 1$$.
