Showing a set function is a premeasure 
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: Consider the collection $S = \{\emptyset, [0, 1], [0, 3], [2, 3]\}$ of subsets of $\mathbb{R}$ and define $\mu(\emptyset) = 0$, $\mu([0, 1]) = 1$, $\mu([0, 3]) = 1$, $\mu([2, 3]) = 1$. Show that  $\mu\colon S\to[0,\infty]$ is a premeasure.

Firstly, we'll show finite additivity. Notice, $\{\emptyset,[0, 1]\}$, $\{\emptyset,[0, 3]\}$,
$\{\emptyset,[2, 3]\}$, $\{\emptyset\}$, $\{[0, 1]\}$, $\{[0, 3]\}$, $\{[2, 3]\}$ are all the finite collections of disjoint sets in $S$ whose unions of each induvial collection if back in $S$.
Further, notice for each induvial collection,
$$\mu(\bigcup\{\emptyset,[0,1]\}) = \mu([0,1]) = 1  = 0 + 1 = \mu(\emptyset) + \mu(\{0,1\})$$
$$\mu(\bigcup\{\emptyset,[0,3]\}) = \mu([0,3]) = 1  = 0 + 1 = \mu(\emptyset) + \mu(\{0,3\})$$
$$\mu(\bigcup\{\emptyset,[2,3]\}) = \mu([2,3]) = 1  = 0 + 1 = \mu(\emptyset) + \mu(\{2,3\})$$
$$\mu(\bigcup\{\emptyset\}) = \mu(\emptyset) = 0  = \mu(\emptyset)$$
$$\mu(\bigcup\{[0,1]\}) = \mu([0,1]) = 1  = \mu([0,1])$$
$$\mu(\bigcup\{[0,3]\}) = \mu([0,3]) = 1  = \mu([0,3])$$
$$\mu(\bigcup\{[2,3]\}) = \mu([2,3]) = 1  = \mu([2,3]).$$
Therefore, the set function $\mu$ is finitely additive.
Now, we'll show countably monotone. For $\emptyset$ any cover will have a sum of measure equal to $0$ or greater than $0$. For $[0,1]$ any cover will have a sum of measure equal to $1$ or greater than $1$. For $[0,3]$ any cover will have a sum of measure equal to $1$ or greater than $1$. For $[2,3]$ any cover will have a sum of measure equal to $1$ or greater than $1$. So, indeed whenever a set $E\in S$ is covered by a countable collection
$\{E_k\}_{k=1}^{\infty}$ of sets in $S$, then the measure of $E$ is less than the measure of the cover.
Therefore, the set function $\mu$ is countably monotone.
As $\emptyset$ belongs to $S$, and $\mu(\emptyset)=0$ by construction, we have that $\mu$ is a premeasure.

My question is, is the above correct?
 A: Your argument is correct. Just small improvements might be useful.

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: Consider the collection $S = \{\emptyset, [0, 1], [0, 3], [2, 3]\}$ of subsets of $\mathbb{R}$ and define $\mu(\emptyset) = 0$, $\mu([0, 1]) = 1$, $\mu([0, 3]) = 1$, $\mu([2, 3]) = 1$. Show that  $\mu\colon S\to[0,\infty]$ is a premeasure.

Firstly, we'll show finite additivity. Notice, $\{\emptyset,[0, 1]\}$, $\{\emptyset,[0, 3]\}$ and $\{\emptyset,[2, 3]\}$, $\{\emptyset\}$, $\{[0, 1]\}$, $\{[0, 3]\}$, $\{[2, 3]\}$ are essentially all the finite collections of disjoint sets in $S$ whose unions of each induvial collection if back in $S$. (We could add additional "empty sets" to those collections, but they would not change the finite additivity, because $\mu(\emptyset)=0$ ).
For collections with a single set (that is  $\{\emptyset\}$, $\{[0, 1]\}$, $\{[0, 3]\}$, $\{[2, 3]\}$ ), it is immediate the finite additivity.
Since $\mu(\emptyset)=0$, it is also immediate the finite additivity for $\{\emptyset,[0, 1]\}$, $\{\emptyset,[0, 3]\}$ and $\{\emptyset,[2, 3]\}$.
Now, we'll show  $\mu$ is countably monotone. Let $E \in S$.
If $E=\emptyset$, any countable cover will have a sum of measure equal or greater to $0$.
If $E= [0,1]$,   any countable cover will have a sum of measure equal  or greater to $1$.
For $E=[0,3]$ any countable cover will have a sum of measure equal  or greater to $1$.
For $E=[2,3]$ any countable cover will have a sum of measure equal  or greater to $1$.
So, indeed whenever a set $E\in S$ is covered by a countable collection
$\{E_k\}_{k=1}^{\infty}$ of sets in $S$, then the measure of $E$ is less than the sum of  measures of the elements in the cover.
Therefore, the set function $\mu$ is countably monotone.
Finally,  $\emptyset$ belongs to $S$, and $\mu(\emptyset)=0$ by the definition of $\mu$. So, we have that $\mu$ is a premeasure.
Remark: For finite additivity, if you want to work case by case, it would be
$$\mu(\bigcup\{\emptyset,[0,1]\}) = \mu([0,1]) = 1  = 0 + 1 = \mu(\emptyset) + \mu([0,1])$$
$$\mu(\bigcup\{\emptyset,[0,3]\}) = \mu([0,3]) = 1  = 0 + 1 = \mu(\emptyset) + \mu([0,3])$$
$$\mu(\bigcup\{\emptyset,[2,3]\}) = \mu([2,3]) = 1  = 0 + 1 = \mu(\emptyset) + \mu([2,3])$$
$$\mu(\bigcup\{\emptyset\}) = \mu(\emptyset) = 0  = \mu(\emptyset)$$
$$\mu(\bigcup\{[0,1]\}) = \mu([0,1]) = 1  = \mu([0,1])$$
$$\mu(\bigcup\{[0,3]\}) = \mu([0,3]) = 1  = \mu([0,3])$$
$$\mu(\bigcup\{[2,3]\}) = \mu([2,3]) = 1  = \mu([2,3]).$$
