Show that monotonicity directly follows countable additivity In these notes http://math.gmu.edu/~dwalnut/teach/Math776/Spring11/776s11lec03_notes.pdf it says countable additivity implies monotonicity of measure
I want to show that $\mu(\bigcup_{k = 1}^\infty E_k) = \sum\limits_{k=1}^\infty \mu(E_k)$ implies that $A \subseteq B$ then $\mu(A) \leq \mu(B)$
This is my solution can someone check if it is right?
Ok so obviously $E_1 \subseteq \bigcup_{k = 1}^\infty E_k$
And we know that $\mu(E_1) + \mu(E_2) + \ldots +  \ldots = \mu(\bigcup_{k = 1}^\infty E_k)$
Then take out all these $\mu(E_k), k \geq 2$  guys
$\mu(E_1) \leq \mu(\bigcup_{k = 1}^\infty E_k)$ 
QED!
Is this proof correct? Can someone offer/direct me to an alternative proof for me to compare? Thanks
 A: The problem with your proof is that you want to show monotonicity for arbitrary sets, not just an instantiated union $\bigcup_{n \in \mathbb{N}} E_n$. Instead, I think that you want to use the countable additivity condition since is true for all disjoint unions, rather than first defining an arbitrary countable union.
Instead:  Let $A \subseteq B$ be  a proper subset. Then $B=A \bigcup (B-A)$.
But then $A \bigcap (B-A)=\emptyset$ so we can utilize countable additivity.
If $B=A$ we are done, since $m(A)=m(B)$.
So wlog let $A \subset B$ be a proper subset. By countable additivity:
Then $m(B)=m(A \bigcup (B-A))=m(A)+m(B-A)$. Since $m(B-A) \geq 0$, it follows that $m(A) \leq m(B)$.
Your proof is not non-salvageable, you would just have to say more about why your proof applies to sets in general. 
edit 
First we need a small lemma: $m(\emptyset)=0$. We first note that measure is nonnegative.
Then: let $A_i=\emptyset$ for all $i \in \mathbb{N}$. Then 
$$m(\emptyset)=m\left(\bigcup_{i \in \mathbb{N}} A_i\right)=\sum_{i \in \mathbb{N}} m(A_i)=m(\emptyset)+m(\emptyset)+m(\emptyset)+...$$ If $m(\emptyset) \neq 0$, then the sum will diverge. We conclude that $m(\emptyset)=0$.
now:countable additivity $\implies$ finite additivity.
Let $\bigcup_{1}^{n} E_n$ be a finite union of disjoint sets. Define $E_{k}=\emptyset$ for all $k>n$.
Then $$m(\bigcup_{n \in \mathbb{N}} E_n)= \sum_{n \in \mathbb{N}} m(E_n)=\sum_{1}^{n}m(E_n)+\sum_{n+1}^{\infty} m(E_n)=\sum_{1}^{n}m(E_n)+m(\emptyset)=\sum_{1}^{n}m(E_n)$$
