$\sigma$-algebra + measure implies premeasure 
Definition: By a measure $\mu$ on a measurable space
$\left(X, \mathcal{M}\right)$ we mean an extended real-valued
nonnegative set function $\mu\colon\mathcal{M}\to[0, \infty]$ for which $\mu(\emptyset) = 0$ and which is countably additive.


Definition: Let $S$ be a collection of subsets of a set $X$ and $\mu\colon S\to[0, \infty]$ a set function. Then $\mu$ is called a premeasure provided $\mu$ is both finitely additive and countably monotone and, if $\emptyset$ belongs to $S$, then $\mu(\emptyset) = 0$.


Question: Show that if a set function on a $\sigma$-algebra is a measure then it's a premeasure.

Let $\mu\colon\mathcal{M}\to[0,\infty]$ be a measure on a $\sigma$-algebra $\mathcal{M}$.
Firstly, as $\mu$ is a measure, by definition it follows that $\mu(\emptyset)=0$.
Finite additivity immediately follows from countable additivity by setting $E_k = \emptyset$, so that
$μ(E_k) = 0$, for $k > n$.
For countable monotonicity, let $\{E_k\}_{k=1}^{\infty}$ be any countable collection of measurable that covers a measurable set $E$. Define $F_1 = E_1$ and then define
$$F_n = E_k\setminus\left[\bigcup_{i=1}^{k-1}E_i\right]\text{ for } k\geq 2.$$
Immediately, it's seen that
$$\{F_k\}_{k=1}^{\infty}\text{ is disjoint, } \bigcup_{k=1}^{\infty}F_k = \bigcup_{k=1}^{\infty}E_k\text{ and } F_k\subseteq E_k\text{ for every } k.$$
From countable additivity and monotonicity of $\mu$ being measure,
$$\mu(E)\leq\mu\left(\bigcup_{k=1}^{\infty}E_k\right)=\mu\left(\bigcup_{k=1}^{\infty}F_k\right) = \sum_{k=1}^{\infty}\mu(F_k)\leq\sum_{k=1}^{\infty}\mu(E_k).$$

Is this correct? Also, I'm not sure the fact that $\mathcal{M}$ is a $\sigma$-algebra comes in?
 A: Your proof is correct. It can be made a little shorter.

Definition: By a measure $\mu$ on a measurable space
$\left(X, \mathcal{M}\right)$ we mean an extended real-valued
nonnegative set function $\mu\colon\mathcal{M}\to[0, \infty]$ for which $\mu(\emptyset) = 0$ and which is countably additive.


Definition: Let $S$ be a collection of subsets of a set $X$ and $\mu\colon S\to[0, \infty]$ a set function. Then $\mu$ is called a premeasure provided $\mu$ is both finitely additive and countably monotone and, if $\emptyset$ belongs to $S$, then $\mu(\emptyset) = 0$.


Question: Show that if a set function on a $\sigma$-algebra is a measure then it's a premeasure.

Proof: Let $\mu\colon\mathcal{M}\to[0,\infty]$ be a measure on a $\sigma$-algebra $\mathcal{M}$.
Firstly, as $\mu$ is a measure, by definition it follows that $\mu(\emptyset)=0$.
Finite additivity immediately follows from countable additivity by setting $E_k = \emptyset$, so that
$μ(E_k) = 0$, for $k > n$.
For countable monotonicity, let $\{E_k\}_{k=1}^{\infty}$ be any countable collection of measurable that covers a measurable set $E$.
Since $\mathcal{M}$ is a $\sigma$-algebra, we have that $\bigcup_{k=1}^{\infty}E_k \in \mathcal{M}$. Since $\mu$ is a measure defined on $\mathcal{M}$, we have
$$\mu(E)\leq\mu\left(\bigcup_{k=1}^{\infty}E_k\right)\leq\sum_{k=1}^{\infty}\mu(E_k)$$
(you don't need to go through the sets $F_n$).
