Proving result about additivity of Lebesgue outer measure The statement I’m trying to prove is that if $\{I_k \}_{k=1} ^{\infty } $ The statement I’m trying to prove is that if $\{I_k \}_{k=1} ^{\infty } $ is a sequence of disjoint intervals in $\mathbb{R}$  and $E_k \subset I_k $ for each $k $ then $$ \mu ^* (\bigcup_{k=1}^{\infty } E_k )= \sum_{k=1}^{\infty } \mu ^* (E_k).$$
I think I need to use the fact that if $(J_k)_k $ is a sequence of disjoint intervals then $\mu^ * (\cup_k J_k )=\sum_k \ell (J_k)$ where $\ell(J_k) $ is the length of the interval $J_k$.
One of the inequalities is trivial by countable subadditivity of $\mu^* $ but I’m struggling with the reverse inequality.
 A: What you search for can be presented as a corollary of a theorem. Below this theorem is stated and proved. This by means of a lemma that is stated and proved first.

Lemma:
If $\varphi$ is an outer measure on
set $\Omega$ and $I_{1},\dots,I_{n}$ are disjoint $\varphi$-measurable
subsets of $\Omega$ then: $$\varphi\left(E\cap\left(\bigcup_{k=1}^{n}I_{k}\right)\right)=\sum_{k=1}^{n}\varphi\left(E\cap I_{k}\right)\text{ for each }E\subseteq\Omega$$
Proof of lemma:
For this we use induction on $n$. Let $I:=\bigcup_{k=1}^{n-1}I_{k}$.
With induction we find: $$\varphi\left(E\cap\left(I\cup I_{n}\right)\right)=\varphi\left(E\cap\left(I\cup I_{n}\right)\cap I_{n}^{c}\right)+\varphi\left(E\cap\left(I\cup I_{n}\right)\cap I_{n}\right)=$$$$\varphi\left(E\cap I\right)+\varphi\left(E\cap I_{n}\right)=\sum_{k=1}^{n}\varphi\left(E\cap I_{k}\right)$$

Theorem:
If $\varphi$ is an outer measure
on set $\Omega$ and $I_{1},I_{2},\dots$ are disjoint $\varphi$-measurable
subsets of $\Omega$ then: $$\varphi\left(E\cap\left(\bigcup_{n=1}^{\infty}I_{n}\right)\right)=\sum_{n=1}^{\infty}\varphi\left(E\cap I_{n}\right)\text{ for each }E\subseteq\Omega$$
Proof of theorem:
Let $I:=\bigcup_{n=1}^{\infty}I_{n}$.
Then for every $n$: $$\varphi\left(E\cap I\right)\geq\varphi\left(E\cap\left(\bigcup_{k=1}^{n}I_{k}\right)\right)=\sum_{k=1}^{n}\varphi\left(E\cap I_{k}\right)$$
according to the lemma above, so that also: $$\varphi\left(E\cap I\right)\geq\sum_{n=1}^{\infty}\varphi\left(E\cap I_{n}\right)$$ Next to that we have: $$\varphi\left(E\cap I\right)\leq\sum_{n=1}^{\infty}\varphi\left(E\cap I_{n}\right)$$ here by countable subadditivity of $\varphi$, so there is equality.

Corollary:
If $\varphi$ is an outer measure on
set $\Omega$ and $I_{1},I_{2}\dots$ are disjoint $\varphi$-measurable
subsets of $\Omega$ then for sets $E_n\subseteq I_n$ we have:$$\varphi\left(\bigcup_{n=1}^{\infty}E_{n}\right)=\sum_{n=1}^{\infty}\varphi\left(E_n\right)$$
Proof of corollary:
Substituting $E=\bigcup_{n=1}^{\infty}E_n$ the theorem gives immediately the desired result.
