Proving a particular function is a measure Statement
Let $(X,\Sigma)$ be a measurable space. Let a function of sets $\mu:\Sigma \to \mathbb R_{\geq 0}$ that satisfies:


*

*$A, B \in \Sigma \space \wedge \space A \cap B = \emptyset \space \implies \mu(A \cup B)=\mu(A) + \mu(B)$

*$A_n \in \Sigma \wedge A_n \searrow \emptyset \implies \lim_{n \to \infty} \mu(A_n)=0$


Prove that $\mu$ is a measure.
The attempt at a solution.
In order to prove that $\mu$ is a measure, I have to show it satisfies three properties:
i. $\mu(\emptyset)=0$
ii. $\mu(E) \geq 0$ for all $E \in \Sigma$
iii. If $\{E_i\}_{i \in \mathbb N}$ is a sequence of pairwise disjoint sets, then $\mu(\bigcup_{i \in \mathbb N} E_i)=\sum_{i \in \mathbb N} \mu(E_i)$
I could prove i., since $\emptyset \in \Sigma$ and $\emptyset \cap \emptyset=\emptyset$, by hypothesis we have $\mu(\emptyset)=\mu(\emptyset \cup \emptyset)=\mu(\emptyset)+\mu(\emptyset) \implies \mu(\emptyset)=0$
I don't know how to show ii. and iii., I would appreciate any help/suggestion.
 A: To prove iii try to define the sets $B_1=E_1$, $B_2=E_2 \backslash E_1$, ... $B_n=E_n \backslash E_{n-1}$...
In this way the sets are disjoint and you can use property 1, the union of the $B_i$'s is the same of the union of the $E_i$'s.
Also, the sequence is shrinking, so you will be able to use property 2 too and prove the result.
ii follows from $\mu: \Sigma \rightarrow R_+$
A: Property ii. follows from the fact that $\mu: \Sigma:\mathbb R_+$. Thus $\mu(E)\ge 0$, for all $E\in\Sigma$.
In order to show property iii. first prove inductively, using 1., that if $E_i\cap E_j=\varnothing$, for $i\ne j$, then
$$
\mu\Big(\bigcup_{i=1}^n E_i\Big)=\sum_{i=1}^n\mu(E_i)\le \sum_{i=1}^\infty\mu(E_i).
$$
Then let $E=\bigcup_{i=1}^\infty E_i$ and $F_n=\bigcup_{i=n+1}^\infty E_i$. 
Clearly
$F_n\cup \Big(\bigcup_{i=1}^n E_i\Big)=E$, $F_n\cap \Big(\bigcup_{i=1}^n E_i\Big)=\varnothing$, 
$$
\mu(E)=\mu(F_n)+\mu\Big(\bigcup_{i=1}^n E_i\Big), \tag{1}
$$
and
$$
F_n\searrow\varnothing,
$$
and thus $\mu(F_n)\to 0$, as a consequence of 2.
Letting $n\to\infty$ in $(1)$ we obtain by virtue of $(1)$ that
$$
\mu(E)=\mu(F_n)+\mu\Big(\bigcup_{i=1}^n E_i\Big)\to 0+\sum_{i=1}^\infty \mu(E_i).
$$
