Compact interval $I=[a,b]\subseteq \mathbb{R}$ is an $\mathfrak{L}$-measurable set.

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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?

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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.