On countable additivity of outer measure If $u^*$ is an outer measure on a set $X$ and if $u^*(A\cup B)=u^*(A)+u^*(B)$, for disjoint $A$ and $B$, then how can I show that $u^*$ is actually countably additive. Any hint to proceed.
 A: Let us present a direct general proof. This proof works even if $u^*$ is not induced by a pre-measure.  
Let $u^*$ be  an outer measure on a set $X$ such that, $u^*(A\cup B)=u^*(A)+u^*(B)$, for disjoint $A$ and $B$. 
For any $\{E_i\}_{i\in \mathbb{N}}$ family of disjoint sets, we have, by induction, for any $n\in \mathbb{N}$,   
$$ u^*\left ( \bigcup_{i=0}^n E_i  \right)=\sum_{i=0}^n u^*(E_i)  $$
On the other hand, for all  $n\in \mathbb{N}$,   $\bigcup_{i=0}^n E_i \subseteq  \bigcup_{i=0}^\infty E_i  $, so we have 
$$ u^*\left (\bigcup_{i=0}^n E_i \right )\leqslant  u^*\left (\bigcup_{i=0}^\infty E_i \right )  $$ 
So, we have 
$$ \sum_{i=0}^n u^*(E_i)= u^*\left ( \bigcup_{i=0}^n E_i  \right) \leqslant  u^*\left (\bigcup_{i=0}^\infty E_i \right ) $$
So, 
$$ \sum_{i=0}^\infty u^*(E_i) = \lim_{n \to \infty}u^*\left ( \bigcup_{i=0}^n E_i  \right) \leqslant  u^*\left (\bigcup_{i=0}^\infty E_i \right )  $$
Now, by the $\sigma$-subadditivity, we have: 
$$ u^*\left ( \bigcup_{i=0}^\infty E_i  \right) \leqslant \sum_{i=0}^\infty u^*(E_i) = \lim_{n \to \infty}u^*\left ( \bigcup_{i=0}^n E_i  \right) \leqslant  u^*\left (\bigcup_{i=0}^\infty E_i \right )  $$
So we have 
$$ u^*\left ( \bigcup_{i=0}^\infty E_i  \right) = \sum_{i=0}^\infty u^*(E_i)   $$
So $u^*$ is countably additive. 
Remark 1: If an outer measure is not not induced by a pre-measure, it may not be continuous from below, and that is why,  to keep the proof above general, we avoided using continuity from below.
Of course, AFTER proving that $u^*$ is $\sigma$-additive, we know that $u^*$ is continuous from below. 
Remark 2: Example of an outer measure (not induced by a pre-measure) that is not continuous from below. 
Let $X=\mathbb{N}$. For any $S \subseteq \mathbb{N}$, if $S\neq \emptyset$ and  $S\neq \mathbb{N}$, let us define $\mu^*(S)=1$, and let us define $\mu^*(\emptyset)=0$ and $\mu^*(\mathbb{N})=2$. It is easy to check that $\mu^*$ is an outer measure. 
Now, for all $n\in \mathbb{N}$, let $I_n=\{ x \in \mathbb{N} : x\leqslant n\}$. It is easy to see that $\{I_n\}_{n\in \mathbb{N}}$ is an increasing family of sets and $\mathbb{N}=\bigcup_{n\in \mathbb{N}}I_n$. However:
$$ \lim_{n \to \infty }\mu^*(I_n) = 1 \neq 2 = \mu^*(\mathbb{N}) $$
So, $\mu^*$ is not continuous from below. 
