Why is the inner measure smaller than the outer measure? If we define the outer measure of a bounded subset $S \subset [a,b]\subset\mathbb{R}$ as
$$
m^* (S) = \inf \sum_{k \in \mathbb{N}} |I_k|, \qquad S\subset \bigcup_{k\in \mathbb{N}} I_k, \quad I_k \quad\mbox{open}
$$
and the inner measure se $m_* (S) = (b-a) - m^* ([a,b]-S)$, is there some simple way to show that 
$$
m^* (S) \geq m_* (S) 
$$
I tried to get a contradiction (I suspect this is the strategy to follow) by assuming $m^* (S) < m_* (S)$, but to my surprise I get only
$$
m^*(S \cup ([a,b]-S)) > m^* (S) + m^*([a,b]-S)
$$ 
which is consistent with the subaditivity of the outer measure.
 A: (0).Equivalently we want to show that $$E\subset [a,b]]\implies m^*(E)+m^*([a,b]\backslash E)\geq b-a.$$(1). Prove that if $F$ is a finite set of open intervals and $\cup F\supset [a,b]$ then $$m(\cup F)\geq \sum_{f\in F}|f|\geq b-a.$$The second inequality is strict but we won't need that........... (2) The interval$[a,b]$ is compact...... (3). With $E\subset [a,b]$, let A and B be open sets with $E\subset A$ and $E^c\subset B$, where $E^c=[a,b]\backslash E$. Now $A=\cup A^*$  and $B=\cup B^*$ where $A^*$ and $B^*$ are families of pairwise-disjoint open intervals, and  $[a,b]\subset (A\cup B)=\ \cup (A^*\cup B^*).$ By (2) there exists a finite $F\subset (A^*\cup B^*)$ such that $[a,b[\subset \cup F$. Therefore by (1)  $$b-a\leq\sum_{f\in F}|f|=\sum_{f\in F\cap A^*}|f|+\sum_{f\in F\cap B^*}|f|\leq m(A)+m(B).$$ For any $e>0$ we can have $m(A)\leq m^*(E)+e $ and $m(B)\leq m^*(E^c)+e$, so $$b-a\leq m^*(E)+m^*(E^c)+2e \text { for any }e>0.$$Pedantic digression:The usual convention is that the sum of an empty set of numbers is $0$. For example $F\cap A^*$ could be empty.
