How can I prove that $\mu$ is a measure? (finite additivity and infinite subadditivity) Let $\Omega$ be a set and $\mathcal{A}$ the $\sigma$-algebra to $\Omega$. Let $\mu : \mathcal{A} \rightarrow [0,\infty]$ be a function with $\mu(\emptyset)=0$, and for sets $A, B \in \mathcal{A}, A \cap B = \emptyset$ let \begin{equation} \mu(A \cup B)=\mu(A)+\mu(B). \end{equation}
Also let $\mu$ be $\sigma$-subadditiv for disjoint sets, that means for $(A_{k})_{k\in \mathbb{N}} \subset \mathcal{A}, A_{i} \cap A_{j}=\emptyset, i\neq j $ let \begin{equation} \mu(\bigcup\limits_{k \in \mathbb{N}}A_{k})\le\sum\limits_{k \in \mathbb{N}}\mu(A_{k}). \end{equation} Prove that $\mu$ is a measure.
 A: To show that $\mu$ is a measure, it is enough to show that if $\{A_n\}_{n=1}^{\infty}$ is a sequence of pairwise disjoint sets, then
$$ \mu\Big(\bigcup_{n=1}^{\infty}A_n\Big)=\sum_{n=1}^{\infty}\mu(A_n).$$
At the moment, we're only guaranteed to have $\leq$ in the line above. So suppose for the sake of contradiction that there is a sequence of pairwise disjoint sets $\{A_n\}_{n=1}^{\infty}$ for which
$$ \mu\Big(\bigcup_{n=1}^{\infty}A_n\Big)<\sum_{n=1}^{\infty}\mu(A_n).$$
Then since $\sum_{n=1}^{\infty}\mu(A_n)=\lim_{N\to\infty}\sum_{n=1}^N\mu(A_n)$, it follows that there is a positive integer $N$ such that
$$ \mu\Big(\bigcup_{n=1}^{\infty}A_n\Big)<\sum_{n=1}^{N}\mu(A_n).$$
However, we know that $\cup_{n=1}^NA_n\subset\cup_{n=1}^{\infty}A_n$, and the finite additivity condition implies that if $A\subset B$ then $\mu(A)\leq \mu(B)$, hence we have
$$ \mu\Big(\bigcup_{n=1}^NA_n\Big)\leq \mu\Big(\bigcup_{n=1}^{\infty}A_n\Big)<\sum_{n=1}^{N}\mu(A_n)$$
which contradicts finite additivity. So equality must hold, and therefore $\mu$ is a measure.
A: Based on the additivity rule: $$\mu(A\cup B)=\mu(A)+\mu(B)$$ for disjoint measurable sets $A,B$ you can prove by induction finite additivity:
$$\mu(A_1\cup\cdots\cup A_n)=\mu(A_1)+\cdots+\mu(A_n)$$ for disjoint measurable sets $A_1,\dots,A_n$.
If $A_1,A_2,\dots$ are disjoint measurable sets and $A=\bigcup_{i=1}^{\infty}A_i$ then:$$A=A_1\cup\cdots\cup A_{n-1}\cup(\bigcup_{i=n}^{\infty}A_i)$$
This is a finite disjoint union of measurable sets so:$$\mu(A)=\mu(A_1)+\cdots+\mu(A_{n-1})+\mu(\bigcup_{i=n}^{\infty}A_i))\geq\mu(A_1)+\cdots+\mu(A_{n-1})$$
This is true for every $n\in\mathbb N$ so that also: $$\mu(A)\geq\sum_{i=1}^{\infty}\mu(A_i)$$
We allready know that $$\mu(A)\leq\sum_{i=1}^{\infty}\mu(A_i)$$ is also true, so end up with:$$\mu(A)=\sum_{i=1}^{\infty}\mu(A_i)$$
