Question on Lebesgue measure Say a set $E$ is $L$-measurable if for all bounded open intervals $(a,b)$ we have $b−a=m^*((a,b)\cap E)+ m^*(E\cap(a,b)c)$. How do we show that any $L$-measurable set is Lebesgue measurable?
Could someone please help me with this. Clear guidelines and help are highly appreciated.
Thank you in advance!
 A: to prove that E is lebesgue measurable we need to show that for all
 $S \subseteq {R} $
 \begin{align*}
m^*(S) = m^*(S \cap E) + m^*(S \cap E^c),
\end{align*}
 case 1: if $S\in R_{leb}$
then it can be expressed as a union of bounded intervals.
let's assume that this union is pairewise disjoint (wlog).
then :$m^*(S)=m^*(\cup I_{k})= \sum m^*(I_{k})= \sum m*((I_{k} \cap E)\cup(I_{k} \cap E^c))$
and since $E$ is $ L-measurable $ then :
 $m*((I_{k} \cap E)\cup(I_{k} \cap E^c))=m^*((I_{k} \cap E))+m^*((I_{k} \cap E^c)$
 therefore: 
 $m^*(S)=\sum m*(I_{k} \cap E)+\sum m^*(I_{k} \cap E^c) \geq  m^*(\cup I_{k}\cap E)+ m^*(\cup I_{k}\cap E^c)=m^*(S \cap E) + m^*(S \cap E^c).$
case 2:  if $S\notin R_{leb} $ 
then: by definition of the outer measure for all $\epsilon>0$: \*
 there exist a covering {$b_{i}$} of S in $R_{leb}$ such that:
 $m^*(S) + \epsilon \geq \sum m^*(bi)$
 on the other hand since $E$ is $L-measurable$:
 \begin{align*}
 \sum m^*(b_{i}) \geq \sum( m^*(b_{i} \cap E)+ m^*(b_{i} \cap E^c))=\sum m^*(b_{i} \cap E)+ \sum m^*(b_{i} \cap E^c) \geq m^*(\cup b_{i} \cap E) +m^*(\cup b_{i} \cap E^c)\end{align*}
 since ${b_{i}}$ is a covering of $S$ then :
$ m^*(\cup b_{i} \cap E) \geq m^*(S \cap E)$ and $m^*(\cup b_{i} \cap E^c) \geq m^*(S \cap E^c)$
consequently : $m^*(S) \geq m^*(S \cap E) + m^*(S \cap E^c)$
and finally  $m^*(S) = m^*(S \cap E) + m^*(S \cap E^c)$ as( $m^*(S) \leq m^*(S \cap E) + m^*(S \cap E^c)$ trivial by the properties of the outer measure)
A: I think something is wrong in your definition. In particular, you want the RHS to $= m_*((a,b)) = b-a.$ 
Suppose your set $E$ satisfies your property
$b-a = m_*((a,b) \cap E) + m_*((a,b) \cap E^c)$, where I assume $m_*$ denotes outer measure.
You now have to show that for all $A \subset \mathbb{R}$, 
\begin{align*}
m_*(A) = m_*(A \cap E) + m_*(A \cap E^c),
\end{align*}
or equivalently for any $E$ and $\epsilon > 0$, there is an open set $O \supset E$ such that
$$m_*(O \setminus E) < \epsilon.$$
The idea will then be to take any set, approximate using a definition of outer measure:
$$m_*(F) = \inf\{ \sum_{i=1}^\infty |B_i| : B_i = B(x_i,r), r > 0\}.$$
(the $B_i$ are open balls of positive radius -- this is equivalent to covering with half-open intervals or whatever). You might even have to use an $\epsilon 2^{-j}$ argument [The trouble here is the $A \subset \mathbb{R}$ is arbitrary, so it is not true that $m_*(O\setminus A)$ can be made arbitrarily small.] 
But no one wants to do that. I think there's a better way. Note that 1) the family $L$ of $L$-measurable sets is an algebra, since it is closed under complements by symmetry and closed under finite unions by symbol-pushing (check this). It also contains the empty set. I claim the family $L$ is a monotone class, i.e. it is closed under countable unions and intersections. Now, show that it is a monotone class, i.e. show that if $E_i \subset E_{i+1}$, $F_j \supset F_{j+1}$, all $F_j, E_i$ are measurable, then $\bigcup_{i=1}^\infty E_i$  and $\bigcap_{j=1}^\infty F_j$ are in the sigma-algebra. 
If the reader is unfamiliar with the monotone class theorem, once we establish that $L$ is a monotone class, it immediately follows that $L$ is actually a $\sigma$-algebra. Since $L$ in particular contains all open intervals, it contains the Borel-sigma algebra, so all $L$ sets are measurable. 
