Are these two definitions of positive measure equivalent? Let's consider the two following definitions of a positive measure :
D1 : Let $\mu$ be an application from a $\sigma$-algebra $T$ to $\overline{\mathbb{R}^+}$.
$\mu$ is called a (positive) measure if and only if :
1)$\mu(\emptyset)=0$
2)for all countable collections $(A_n)_{n\in\mathbb{N}}$ of pairwise disjoint sets of T , $$\mu(\bigcup_{n \in \mathbb{N}}A_n)=\sum_{n \in \mathbb{N}}\mu(A_n)$$
D2 : Let $\mu$ be an application from a $\sigma$-algebra $T$ to $\overline{\mathbb{R}^+}$.
$\mu$ is called a (positive) measure if and only if :

*

*$\mu(\emptyset)=0$
2)$\mu$ is additive , which means for all disjoint pairs $A,B$ of T , $\mu(A\cup B)=\mu(A)+\mu(B)$


*for all sequence $(A_n)_{n \in \mathbb{N}}$ of $T$ such that $A_{n+1}\subset{A_n} ,\forall n \in \mathbb{N}$ and $\mu(A_0)<\infty$ , we have $$\mu(\bigcap_{n \in \mathbb{N}} A_n)= \lim_{n \to \infty} \mu(A_n)$$
EDIT:
I add a 4th condition :
4)for all sequence $(A_n)_{n \in \mathbb{N}}$ of $T$ such that $A_{n}\subset A_{n+1} ,\forall n \in \mathbb{N}$ , we have $$\mu(\bigcup_{n \in \mathbb{N}} A_n)= \lim_{n \to \infty} \mu(A_n)$$
I think it can show that $(D_1)\implies(D_2)$ , but does $(D_2)\implies (D_1)$ holds ?
Is there a proof by equivalence ?
 A: Item 1. $D2$ without condition $4$ does not imply  $D1$.
Consider $\mu$ defined on $\mathcal{P}(\Bbb N)$ by
$ \mu(E) = \sum_{n \in E} \frac{1}{2^n} $,  if $E$ is finite, and
$ \mu(E)  = +\infty$ , if  $E$ is infinite.
It is easy to see that $\mu$ satisfies conditions $1$, $2$ and $3$  of $D2$,   but not condition $2$ of $D1$. In fact, $\Bbb N =\bigcup_{n \in \Bbb N}\{n\}$ and
$$ \mu(\Bbb N ) = +\infty \ne 2 = \sum_{n \in \Bbb N}\mu(\{n\}) $$
Item 2. On the other hand, $D2$ with only conditions $1$, $2$ and $4$  (no need for condition $3$) implies $D1$.
Proof:  Condition $1$ is the same in $D1$ and $D2$. So, it remains to be proved that conditions $1$, $2$ and $4$  of $D2$ implies condition $2$ of $D1$.
Given any countable collections $(A_n)_{n\in\mathbb{N}}$ of pairwise disjoint sets of T, for all $k \in \mathbb{N}$, let $B_k = \bigcup_{n=0}^k A_n$. Then, we have that $B_{k}\subset B_{k+1} ,\forall k \in \mathbb{N}$ and
$$ \bigcup_{k \in \mathbb{N}} B_k = \bigcup_{n \in \mathbb{N}} A_n$$
So, using condition $4$ and $2$ of $D2$, we have
$$\mu(\bigcup_{n \in \mathbb{N}} A_n) = \bigcup_{k \in \mathbb{N}} B_k= \lim_{k \to \infty} \mu(B_k) = \lim_{k \to \infty} \sum_{n=0}^k\mu(A_n)= \sum_{n \in \mathbb{N}}\mu(A_n)$$
So, condition $2$ of $D1$ is satisfied. $\square$
