I want to prove a proposition, that tells us a way to generate outer measures. I will first start with the definition of
Definition (outer measure) Let $X$ be a set. An outer measure is a function $\mu^*$ defined on the collection of all subsets of $X$ such that
(i) $\mu^*(\emptyset)=0$
(ii) if $A \subseteq B \Rightarrow \mu^*(A)\leq \mu^*(B)$
(iii) $\mu^*(\cup_{i=1}^{\infty} A_i)\leq \sum_{i=1}^{\infty} \mu^*(A_i)$, where $A_i$ are arbitrary subsets of $X$.
Proposition
Suppose $\mathcal{C}$ is a collection of subsets of $X$ such that (1) $\emptyset \in \mathcal{C}$
(2) there exist $D_1,D_2,... \in \mathcal{C}$ such that $X=\cup_{i=1}^{\infty}D_i$.
Further, suppose $l:\mathcal{C}\rightarrow [0,\infty]$ with $l(\emptyset)=0$.
Define $\mu^*(E):=\inf\{\sum_{i=1}^{\infty} l(A_i): A_i \in \mathcal{C} \text{ for all i and } E\subseteq \cup_{i=1}^{\infty} A_i\} $
Then $\mu^*$ is an outer measure.
Proof: (i) holds by definition of $l$. (ii) holds as a property of the infimum, i.e. if $A \subset B \Rightarrow infA \leq infB$ Now we came to (iii): To get a better feeling of what is going on, I wanted to first show a weaker condition than (iii). I want to show that, for $A,B \subseteq X$ we get: $\mu^*(A \cup B)\leq \mu^*(A) +\mu^*(B)$.
For this prove I will use the \epsilon-definiton of the infimum, i.e. I is the infimum of some set M, iff
- I is a lower bound of M
- $\forall \epsilon>0 \exists m \in M: I+\epsilon>m$
So, lets start: Since $A \subseteq X$ we can find a cover $A_i$ such that $A\subseteq \cup_{i=1}^{\infty} A_i$. By the same argument,$B\subseteq \cup_{i=1}^{\infty} B_i$.
Since $\mu^*(A)$ is the infimum of the set of sums of the form $\sum_{i=1}^{\infty} A_i $, we can find a $\epsilon/2$ such that $\mu^*(A)+\frac{\epsilon}{2}> \sum_{i=1}^{\infty}A_i$
The same applies to $\mu^*(B)$ and we get: $\mu^*(B)+\frac{\epsilon}{2}> \sum_{i=1}^{\infty}B_i$
Now since $A\subseteq \cup_{i=1}^{\infty} A_i, B\subseteq \cup_{i=1}^{\infty} B_i$ we know that
$A \cup B \subseteq \bigl(\cup_{i=1}^{\infty} A_i \bigr) \bigcup \bigl(\cup_{i=1}^{\infty} B_i \bigr)$. Thus we can find a covering $\cup_{i=1}^{\infty} (A_i \cup B_i)$ of $A \cup B$.
Now the idea would be to write
$\mu^*(A \cup B)\leq \sum_{i=1}^{\infty}l(A_i \cup B_i)=\sum_{i=1}^{\infty} l(A_i)+\sum_{i=1}^{\infty} l(B_i)\leq \mu^*(A )+ \frac{\epsilon}{2}+\mu^*(B)+ \frac{\epsilon}{2}=\mu^*(A )+ \mu^*(B)+ \epsilon$
And since this wolds for all $\epsilon$ we get the result.
But I do not know how to argue the step $\sum_{i=1}^{\infty}l(A_i \cup B_i)=\sum_{i=1}^{\infty} l(A_i)+\sum_{i=1}^{\infty} l(B_i)$. Since the only thing about $l$ I know is that it maps the empty set to zero. It was never mentioned that it behaves linear or any other property.
You're close, but this isn't quite right:
First, there's no reason why $A_i \cup B_i \in \mathcal{C}$, so $l(A_i \cup B_i)$ may be nonsense, and second, even if it does make sense, we might have $l(A_i\cup B_i) > l(A_i) + l(B_i)$, in which case you couldn't continue as you had.
Instead, what you actually want to write is $$\mu^*(A \cup B)\leq \sum_{i=1}^{\infty} l(A_i)+\sum_{i=1}^{\infty} l(B_i)\leq \mu^*(A )+ \frac{\epsilon}{2}+\mu^*(B)+ \frac{\epsilon}{2}=\mu^*(A )+ \mu^*(B)+ \epsilon$$
If you want the first inequality to be more clear, define the sequence $C_n \in \mathcal{C}$ by $C_{2i-1} = A_i$ and $C_{2i} = B_i$, so as a result $$A\cup B \subseteq \bigcup_{i=1}^\infty A_i \cup \bigcup_{i=1}^\infty B_i = \bigcup_{i=1}^\infty C_i$$ and then by the definition of $\mu^*$, $$\mu^*(A\cup B) \leq \sum_{i=1}^\infty l(C_i) = \sum_{i=1}^\infty l(C_{2i-1}) + \sum_{i=1}^\infty l(C_{2i}) = \sum_{i=1}^\infty l(A_i) + \sum_{i=1}^\infty l(B_i)$$