I have to show that an open interval in the form $(a,b)$, where $a,b \in {\mathbb R}$ and $a < b$ is Lebesgue measurable. I think I'm supposed to show, that the subset $(a,b)$ is Lebesgue measurable, if and only if:
$$m(A) = m(A ∩ S) + m(A ∩ S^c)$$
where $S \subseteq {\mathbb R}^n$ and $S^c$ is the complement of $S$. But how do I actually prove that the open interval $(a,b)$ is Lebesgue measurable?
I assume you are considering sets in $\mathbb R.\ $ Let $A\in \mathcal P (\mathbb R),\ I=(a,b)$ and $\ \epsilon>0.\ $ For convenience, denote both the outer measure and length of intervals by $|\cdot|.$
There is a sequence $(I_n)$ of intervals such that $\bigcup I_n\supseteq A$ and $\sum |I_n|<|A|+\epsilon.\ $ Set $J_n=I\cap I_n;\ J_n'=I^c\cap I_n.\ $ Some of these may be empty, but that's ok.
Then,
$|I_n|=|J_n|+|J_n'|,\ A \cap I \subseteq \bigcup_{n=1}^{\infty} J_n,\ A \cap I^c \subseteq \bigcup_{n=1}^{\infty} J'_n,\ $ and these facts imply that $|A\cap I|+|A\cap I^c|\le |A|+\epsilon.$