Define $m_*(E) = \inf_{E \subset \bigcup_{i=1}^\infty J_i} \sum_{i=1}^\infty \ell(J_i)$ where $\bigcup_{i=1}^\infty J_i$ is an open cover of $E$ (i.e. an outer measure in $\mathbb{R}$). Let $\ell$ denote the length function.
If the $I_i$'s are intervals and $I_i \cap I_j = \emptyset$ $\forall i,j$, prove $m_*( \bigcup_{i=1}^\infty I_i ) = \sum_{i=1}^\infty \ell ( I_i )$.
Attempt:
By the definition of outer measure, $$ m_*(\bigcup_{i=1}^\infty I_i ) \leq \sum_{i=1}^\infty \ell(I_i).$$
To prove the other inequality, note that $$ m_*(\bigcup_{i=1}^NI_i ) \leq m_*(\bigcup_{i=1}^\infty I_i ).$$
At this point, I'm stuck. I would like to assert ** below so that $$ \sum_{i=1}^N\ell(I_i) \underset{**}{\leq} m_*(\bigcup_{i=1}^\infty I_i ) \leq \sum_{i=1}^\infty \ell(I_i)$$ and take $N \to \infty$. However, I think I need countable additivity to claim that $$ m_*(\bigcup_{i=1}^NI_i ) \leq m_*(\bigcup_{i=1}^\infty I_i ) \implies \sum_{i=1}^N\ell(I_i) \leq m_*(\bigcup_{i=1}^\infty I_i ).$$ Is the solution as simple as stating that the inequality ** follows because the $I_i$'s are disjoint?