Proving every compact set on $\mathbb{R}$ is measurable Before starting proof, my knowledge includes followings;
(m represents the measurement function, $\lambda_*$ and $\lambda^*$ represent Lebesgue inner and outer measure)
i) Let $U$ is an $d_E$-open set then $U$ can be written by using disjoint intervals $\{I_n\}$ which are finite or countable sets and   $U= \bigcup_n I_n$ $\Rightarrow$ $m(U)=\sum_n m(I_n)$.
and
ii) Let $U$ and $V$ are $d_E$-open sets on $\mathbb{R}$
$m(U)+m(V)=m(U \cup V) + m(U \cap V)$
iii) $E \subset \mathbb{R}$ is a bounded set. $E$ can be measured iff $\lambda^*(E)=\lambda_*(E)$
So I have started proof by determining my aim for measuring the compact set. So I think that because of measuring only open sets on $\mathbb{R}$ I have to write the compact set into combination of open sets then I can measure the compact set.
Let K is a compact set, U is an arbitrary open set and $K \subset U$. Because of $K \subset U$ we can re-write K as
$K= U \setminus (U\setminus K)$ and we have open and disjoint sets $U$ and $(U \setminus K)$  Now 
$m(K)=m(U \setminus(U \setminus K))=........$
which I got for now. I don't know what am I going to do because I have no information about measuring the set difference. Also I didn't use the information iii) because while determining the Lebesgue inner measure,
$\lambda_*(E)=sup\{ m(K) : K \subset E, K compact\}$ 
I have to calculate a compact set's measure.
Thanks for any help 
 A: In $\Bbb{R}$ every compact set is closed and bounded form Heine-Borel.
So it suffices to show that $K^c$ which is open,is measurable.
We have that in $\Bbb{R}$ every open set can be expressed as a countable union of disjoint open intervals,thus $$K^c=\bigcup_{n=1}^{\infty}(a_n,b_n)$$
So we can show that every open interval $(a_n,b_n)$ is measurable.
A set is measurable on the real line if it can be approximated with respect to the Lebesgue $\text{outer}$ measure by open supersets.
In  other words:
Let $\epsilon>0$.
Take $B=(a,b+\frac{\epsilon}{2}) \supset (a,b)$ and we have that $m^*(B \setminus (a,b))=m^*(b,b+\frac{\epsilon}{2})=m^*(b,b+\frac{\epsilon}{2})=l(b,b+\frac{\epsilon}{2})\leq \epsilon$
where $l$ denotes the length of an inteval and we know that the outer measure of an interval is equal to its length.
So $(a,b)$ is measurable and from this you can deduce that $K^c$ is measurable as a countable union of measurable sets thus $K$ is measurable.
In general take the Lebesgue outer measure $m^*$.
A set $E$ on the real line is $m^*$ measurable if $m^*(A)=m^*(A \cap E)+m^*(A \cap E^c),\forall A \subseteq \Bbb{R}$.$(*)$
In general we consider an outer measure $\mu^*$ and we can construct the sigma algebra of subsets of $\Bbb{R}$ for instance with the property $(*)$.
Take a look at this:
https://math.dartmouth.edu/~ddeford/103_day4.pdf
With the Lebesgue outer measure $m*$ and the above technique we can construct the sigma algebra of $m^*-$measurable sets.
So the restriction of $m^*$ in this sigma algebra is the Lebesgue measure.
The argument i used in the proof just contains properties of the Lebesgue measure on the real line.
So now you can see that a set is measurable iff its complement is measurable from the defintion of a sigma algebra.
