I have just showed that if $\{I_k \}_k $ is a collection of disjoint open intervals of $\mathbb{R}$ that $$ \lambda ^* \left( \bigcup_{k=1}^{\infty }I_k \right) = \sum_{k=1}^{\infty } \ell (I_k ) $$ where $\ell $ just denotes the length of the interval.
Now I am trying to show that if $A \subset \mathbb{R}$ that $$ \lambda^* \left( A \cap \left(\bigcup_{k=1}^{\infty } I_k\right) \right)= \sum_{k=1}^{\infty } \lambda^* (A\cap I_k).$$
Since one direction inequality hold any countable subaddivity of $\lambda ^* $ it is just the other inequality I am having trouble showing. Any pointers?
What you want to show is that if $(I_k)_k$ is a sequence of disjoint real intervals and $\forall k\in\Bbb N\quad E_k\subset I_k,$ then $$\lambda^*(\cup_{k\in\Bbb N}E_k)=\sum_{k\in\Bbb N}\lambda^*(E_k),$$ i.e. $$\inf\sum_{k,n\in\Bbb N}\ell(I_{k,n})=\sum_{k\in\Bbb N}\inf\sum_{n\in\Bbb N}\ell(I_{k,n})$$ where on both sides, the infimum is taken over intervals $I_{k,n}$ such that $E_k\subset\cup_{n\in\Bbb N}I_{k,n}$ and wlog, $I_{k,n}\subset I_k.$ Writen like that, it becomes straight forward.