Show that $\mu^*(\mathcal{U})=\sum_n \ell(I_n).$ Suppose $\mathcal{U}$ is an open set which is the union of a sequence of disjoint open intervals, $I_1,I_2,I_3,\dots$. Show that $$\mu^*(\mathcal{U})=\sum_n \ell(I_n).$$
Proof: Let $\mathcal{U}=\cup_{n=1}^\infty I_n$ be an open set such that $\cap_{n=1}^\infty I_n=\emptyset$ where $I_n$ is an open set for all n. We need to show $$\mu^*(\mathcal{U})=\sum_n \ell(I_n).$$ By subaddivitity, we have $$\mu^*(\mathcal{U}) \leq \sum_n \mu(I_n)=\sum_n \ell(I_n).$$ We need to show that $$\mu^*(\mathcal{U}) \geq \sum_n \ell(I_n).$$ By Caratheodory's Construction,
\begin{equation*}
\begin{aligned}
\mu^*(\mathcal{U}) & =\mu^*\left(\cup_{n=1}^\infty I_n\right) \\
& =\mu^*\left(\cup_{n=1}^\infty I_n\cap I_1 \right)
+ \mu^*\left(\cup_{n=1}^\infty I_n \cap I_1^c\right) \\
& =\mu^*\left( I_1 \right)
+ \mu^*\left(\cup_{n=2}^\infty I_n\right) \\
& =\ell\left( I_1 \right)
+ \mu^*\left(\cup_{n=2}^\infty I_n\right) \\
\end{aligned}
\end{equation*}
If we repeat the process enough times we get that 
\begin{equation*}
\begin{aligned}
\mu^*\left(\mathcal{U}\right) 
& =\sum_{n=1}^i \mu^*\left( I_n \right)
+ \mu^*\left(\cup_{n=i+1}^\infty I_n\right) \\
& =\sum_{n=1}^i \ell\left( I_n \right)
+ \mu^*\left(\cup_{n=i+1}^\infty I_n\right) \\
& \geq \sum_{n=1}^i \ell\left( I_n \right)
\end{aligned}
\end{equation*}
Hence, $$\mu^*(\mathcal{U})=\sum_n \ell(I_n).$$
 A: If $\mathcal{U}$ is the disjoint union of countably many open intervals. Since each $I$ is measurable, and since $\mathcal{M}$ is a $\sigma-\textrm{algebra}$, it is closed under countable union. Then $\mathcal{U}$ is measurable, and hence
 $$m^*(\mathcal{U})=m(\mathcal{U})=m(\bigsqcup_{I \in \mathscr{C}} m(I))=\sum_{I \in \mathscr{C}} l(I)$$
However, we can go ahead and try to prove something much stronger:
Lemma 0 Every open subset of $\mathbb{R}$ is a countable disjoint union of open intervals, and is hence measurable.
First seperate the open subset $U \subset \mathbb{R}$ into components by imposing the equivalence relation $x~y$ iff $(x,y) \in U$. Use this to prove that each component is either an interval or a ray.
hint: define $a= \inf(U)$, $b=\sup(U)$. Let $x \in (a,b)$. Then find an open ball about $x$ so that it is between $a$ and $b$. The result will follow shortly thereafter.
Then utilize the countable basis for $\mathbb{R}$ defined by
$$\mathbb{B}= \{\beta(q, 1/n) \mid q \in \mathbb{Q}, n \in \mathbb{N}\}$$, and use the fact that the rationals are dense in $\mathbb{R}$, and hence there are only countably many disjoint components.
Lemma 1 is due to James Belk:
$L(\mathcal{I})$ denotes the length of the interval.
Lemma 1: Let $S \subseteq \mathbb{R}$. Then 
$m^*(S)=\textrm{inf} \{ m(U) \mid U \textrm{is open}, S \subseteq U\}$.
proof: Let $x$ be the value of the infimum. Clearly $m^*(S) \leq m(U)$ by monotonicity, implying that $m^*(S) \leq x$. For the other side of the inequality: let $\epsilon>0$, and let $\mathscr{C}$ be an open cover of $S$ so that $$\sum_{I \in \mathscr{C}} L(I) \leq m^*(S)+\epsilon$$
Then  $U \bigcup \mathscr{C}$ contains $S$, so
$$x \leq m(U) \leq \sum_{I \in \mathscr{C}} m(I)=\sum_{I \in \mathscr{C}} m(I) \leq m^*(S)+ \epsilon.$$
The result follows.
Lemma 2: Let $(x,\mathcal{M},\mu)$ be a measure space, and define $\mu^{*}:\mathbb{P}(x) \to [0,\infty]$ by 
 $$\mu^{*}(s)= \textrm{inf}\{\mu(E) \mid E \in \mathcal{M}, \, S \subseteq E\}$$. 
 Then for each $S \in \mathbb{P}(x)$, there exists $E \in \mathcal{M}$ so that $S \subseteq E$ and $\mu(E)=\mu^{*}(S)$.
Let $S \in \mathbb{P}(x)$. let $\{E_n\}$ be a sequence of measurable sets in $\mathcal{M}$ so that $S \subseteq E_n$ and $\mu(E_n) \leq \mu^{*}(S)+1/n$ for each $n \in \mathbb{N}$. Let $E= \cap_{n \in \mathbb{N}}E_n$. Then $E \in \mathcal{M}$, $S \subseteq E$, and $\mu(E) \leq \mu^{*}(S)+1/n$, implying that $\mu^{*}(S)=\mu(E)$.
Corollary: For all $S \subseteq \mathbb{R}$, $m^*(S)=\sum_{I \in \mathscr{C}} l(I)$ for some  disjoint union of open intervals.
This follows by countable additivity. This result is a little bit stronger, and is pretty cool.
For measures in general, the components proof will suffice to show the measurability of open sets for all measure spaces that have a countable basis.
