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.

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.
