Compact interval $I=[a,b]\subseteq \mathbb{R}$ is an $\mathfrak{L}$-measurable set. Let $\mathfrak{L}$ denote the Lebesgue (outer) measure on $\mathbb{R}$. Show, directly using the definition of $\mathfrak{L}$, that a compact interval $I=[a,b]\subseteq \mathbb{R}$ is a $\mathfrak{L}$-measurable set.
I proved that Lebesgue measure is equal to length of the interval. But i need to prove more. I need to use Carathéodory's criterion. Can you help me?
 A: For any $A\subset\mathbb{R}$, by definition of outer measure, there exists a sequence of open sets $\{I_n\}_{n=1}^\infty$ such that $A\subset \cup_{n=1}^\infty I_n$ and $\sum_{n=1}^\infty m^*(I_n)\leq m^*(A)+\epsilon$. Since $A\cap I\subset \cup_{n=1}^\infty (I_n\cap [a,b])$,
\begin{equation}
m^*(A\cap I)\leq \sum_{n=1}^\infty m^*(I_n\cap [a,b])
\end{equation}
Similarly, $A\cap I^c\subset (\cup_{n=1}^\infty(I_n\cap (-\infty,a)))\cup (\cup_{n=1}^\infty(I_n\cap (b,\infty)))$. Therefore,
\begin{equation}
m^*(A\cap I^c) \leq \sum_{n=1}^\infty (m^*(I_n\cap (-\infty,a)))+m^*(I_n\cap (b,\infty)))
\end{equation}
Combined with the inequality for $m^*(A\cap I)$, it yields
\begin{align}
m^*(A\cap I)+m^*(A\cap I^c) &\leq \sum_{n=1}^\infty (m^*(I_n\cap (-\infty,a))+m^*(I_n\cap (b,\infty))+m^*(I_n\cap [a,b])) \\
&= \sum_{n=1}^\infty m^*(I_n\cap (-\infty,\infty)) = \sum_{n=1}^\infty m^*(I_n) \leq m^*(A)+\epsilon
\end{align}
Take $\epsilon\to 0$, we have $m^*(A\cap I)+m^*(A\cap I^c)\leq m^*(A)$. Using countable subadditivity of outer measure, we obtain the reverse inequality. This verifies the Carathéodory's criterion.
