I am trying to prove that if $\cup_{i=1}^{\infty}I_i$ are disjoint open intervals then $$|\cup_{i=1}^{\infty}I_i|=\sum_{i=1}^{\infty}\mathcal{l}(I_i).$$ Where $||$ is the Lebesgue outer measure.
I know that by countable subadditivity we have $$|\cup_{i=1}^{\infty}I_i|\le\sum_{i=1}^{\infty}\mathcal{l}(I_i).$$
For the other direction consider $\epsilon>0$ then by definition of the infimum there must exist $ \cup_{i=1}^{\infty}I_i \subset \cup_{k=1}^{\infty}J_k$ such that
$$|\cup_{i=1}^{\infty}I_i|+ \epsilon \ge\sum_{k=1}^{\infty}\mathcal{l}(J_k)$$.
Consider the covering $I_i\subset\cup_{k=1}^{\infty}J_k\cap I_i$, then $$\sum_{k=1}^{\infty}\mathcal{l}(J_k\cap I_i)\ge |I_i|.$$
This implies
$$\sum_{i=1}^{\infty}\sum_{k=1}^{\infty}\mathcal{l}(J_k\cap I_i)\ge \sum_{i=1}^{\infty}|I_i|=\sum_{i=1}^{\infty}\mathcal{l}(I_i).$$ Since $$\sum_{k=1}^{\infty}\mathcal{l}(J_k)\ge\sum_{i=1}^{\infty}\sum_{k=1}^{\infty}\mathcal{l}(J_k\cap I_i),$$
we get
$$|\cup_{i=1}^{\infty}I_i|+ \epsilon \ge \sum_{i=1}^{\infty}\mathcal{l}(I_i).$$
Since $\epsilon$ was arbitrary we get
$$|\cup_{i=1}^{\infty}I_i| \ge \sum_{i=1}^{\infty}\mathcal{l}(I_i).$$
I don't have confidence in my proof abilities. Are there any mistakes in my proof?