Outer Measure of a countable union of disjoint intervals If $E=\cup^{\infty}_{n=1} I_n$ is a countable union of pairwise disjoint intervals, I want to show that $m^*(E)=\sum^{\infty}_{n=1} m^*(I_n)$, where $$ m^*(E):=\{\sum^{\infty}_{n=1} Length(J_n): E\subset \cup^{\infty}_{n=1} J_n \} $$ each $J_n$ is a bounded interval, is the outer measure of $E$.
My strategy:
Since $E = \cup^{\infty}_{n=1} I_n$  and because the $I_n$ are disjoint, 
$$m^*(E) = m^*(\cup^{\infty}_{n=1} I_n) = \sum^{\infty}_{n=1} m^*(I_n)= \sum^{\infty}_{n=1} Length(I_n)$$
Provided I know that $m^*(I_n) = Length(I_n)$ for all intervals, does this proof work. I feel as though I am missing something.
 A: I think there is more to say than what you have when you wrote down the list of equalities. 
I will use the fact that if $(I_n)$ and $(J_k)$ are sequences of intervals such that $\bigcup\limits_{n=1}^{\infty} I_n = \bigcup\limits_{k=1}^{\infty} J_k$ and the $(I_n)$ are pairwise disjoint, then $\sum_{n=1}^{\infty} l(I_n) \leq \sum_{k=1}^{\infty} l(J_k).$ (Where $l(I_n)$ means the length of the interval.) A proof of this can be found in Carother's Analysis on page 268.
By assumption $E = \bigcup\limits_{n=1}^{\infty} I_n$, so $m^{*}(E) = m^{*}(\bigcup\limits_{n=1}^{\infty}) \leq \sum_{n=1}^{\infty} l(I_n) = \sum_{n=1}^{\infty} m^{*}(I_n)$.
Now by definition of Outer Measure given $\epsilon > 0 $ there is a sequence of intervals $(J_k)$ such that $E \subseteq \bigcup\limits_{k=1}^{\infty} J_k$ and $\sum_{k=1}^{\infty} l(J_k) \leq m^{*}(E) + \epsilon$. 
Then because the $I_n$'s are pairwise disjoint we have $\sum_{n=1}^{\infty} l(I_n) \leq \sum_{k=1}^{\infty} l(J_k) \leq m^{*}(E) + \epsilon$.
Letting epsilon go to zero we have the other inequality, thus
$\sum_{n=1}^{\infty} l(I_n) = m^{*}(E)$.
