For $A\subset \mathbb{R}$ , prove that : Exist $K\subset A$ compact : $\mu(A\backslash K)< \epsilon$ For $A\subset \mathbb{R}$, $A$ is measurable Lebesgue and $\mu(A)<\infty$ . Prove that $\forall \epsilon>0$ Exist $K\subset A$ compact : $\mu(A\backslash K)< \epsilon$

If $A$ is Lebesgue measurable we have $b^{*}(A)=b_{*}(A)=\mu(A)$, where $b$ denotes a classical one-dimensional Borel measure in $R$ (Notice that a usual completion of $b$ is $\mu$). The relation $b_{*}(A)=\mu(A)$ means that for $n \in N $ there is a Borel subset $B_n$ of $A$ such that $\mu(A \setminus B_n)<1/n$. If we consider a Borel set $\cup_{n \in N}B_n$, it again is a Borel subset of  $A$ and $\mu(A \setminus \cup_{n \in N}B_n)=0$. Since $b$ is inner regular for $\epsilon >0$ there is a compact set $K \subseteq  \cup_{n \in N}B_n$ such that $b(\cup_{n \in N}B_n \setminus K)<\epsilon$. Now it is obviuos that 
$$
\mu(A\setminus K)=\mu((A \setminus \cup_{n \in N}B_n)\cup (\cup_{n \in N}B_n \setminus K))=
$$
$$\mu(A \setminus \cup_{n \in N}B_n)+\mu(\cup_{n \in N}B_n \setminus K)=0+b(\cup_{n \in N}B_n \setminus K)<\epsilon.
$$
