I have been working through the Axler measure theory exercises and came across the following problem that I have been stuck of for quite some time. \begin{array}{l}\text { Let } X=\mathbb{R} \text { and let } \mathcal{S}=\{A \subseteq \mathbb{R}: \text { either } A \text { is countable or } \mathbb{R} \backslash A \text { is countable }\} . \text { Let } f: \mathbb{R} \rightarrow \\ {[0, \infty) \text { be any function, and define } \mu_{f}: S \rightarrow[0, \infty] \text { by }} \\ \qquad \mu_{f}(A)=\sup \left\{\sum_{x \in F} f(x): F \subseteq A \text { is finite }\right\} . \\ \text { Show that } \mu_{F} \text { is a measure on }(\mathbb{R}, \mathcal{S}) \text { . } \end{array}
I know I need to try and prove that:
i. $\mu(\emptyset)=0$
ii. $\mu() \geq 0$ for all $E \in S$
iii. If $\{ _ \}, \in \mathbb{N}$ is a sequence of pairwise disjoint sets, then $\mu\left(\bigcup_{i \in \mathbb{N}} E_{i}\right)=\sum_{i \in \mathbb{N}} \mu\left(E_{i}\right)$.
i. seems obvious to me but im having trouble with ii and iii.
I would appreciate any help/suggestion.
The second one seems also trivial, as the finite sum of non-negative amounts is itself a non-negative amount and $\sup(A)>=a $ $\forall a \in A$ making it non-negative as well, so it all follows from the fact that $f(x)$ is non-negative for all x.
Maybe for the third one it is easier to prove first that what you have is an outer measure and then prove that every set in the $\sigma$-algebra you are given is $\mu$-measurable. I don't have paper to run numbers now, give it a shot and maybe I'll edit this if I can look further into it.
EDIT: We will prove the additivity of the measure. For that, first I need a little lemma
Lemma: let $A, B \in S$. Then $\mu(A\cup B) = \mu(A) + \mu(B)$
This is quite easy. Suppose we have a sequence of finite sets $A_i$ such that $\sum_{x\in A_i}f(x)\rightarrow \mu(A)$, and an analogous for $B$. Then, $C_i=A_i\cup B_i$ is a sequence of finite sets in $A\cup B$ and trivially $$\mu(A\cup B) = \sup\{\sum_{x\in C_i}f(x)\} = \sup\{\sum_{x\in A_i}f(x)+\sum_{x\in B_i}f(x)\} = \mu(A) + \mu(B)$$
With that lemma proven, it is trivial that $g(N) = \mu(\bigcup_{i=1}^NA_i) = \sum_{i=1}^N\mu(A_i)$. Considering the fact that this sequence is non-decreasing, it is trivial that $$\mu(\bigcup_{n\in\mathbb{N}}A_i) = \sup\{g(N) : N\in\mathbb{N}\} = \lim_{N\to\infty}g(N) = \sum_{n\in\mathbb{N}}\mu(A_i)$$ and so $\mu$ is in fact a measure