Isn't $\mathbb{P}$ already a probability measure, so what is there to prove? Follow-up to Probability measure over finite sample space. 
This is a theorem from Casella and Berger's Statistical Inference:

Let $S = \{s_1, \dots, s_n\}$ (sample space) be finite and $p_1,
 \dots, p_n$ be nonnegative for all $i \in \{1, 2, \dots, n\}$ and
  $\sum\limits_{i=1}^{n}p_i = 1$. For any $A \in \mathcal{B}$ (a
  $\sigma$-algebra of subsets of $S$), define $\mathbb{P}: \mathcal{B} \to [0, 1]$ by
  $$\mathbb{P}(A) = \begin{cases}  \sum_{\{i \mid s_i \in A\}}p_i, & A
 \neq \varnothing \\  0, & A = \varnothing\text{.}  \end{cases}$$  Then
  $\mathbb{P}$ is a probability measure. 

I asked if there was a correspondence between $s_i$ and $p_i$ for all $i$, and I was told that $\mathbb{P}(\{s_i\}) = p_i$, as you can see in the link.
I am trying to follow the proof of this in the textbook. It proves $\mathbb{P}(A) \geq 0$, $\mathbb{P}(S) = 1$, and to prove that $\mathbb{P}$ is countably additive, it says to consider pairwise disjoint events $A_1, A_2, \dots, A_k$. We can consider $k < \infty$ since $S$ is finite, sure. [Are some steps skipped here that I don't know of?]
But then the proof goes like $$\mathbb{P}\left(\bigcup_{i=1}^{k}A_i\right) = \sum\limits_{\{j\mid s_j \in \bigcup_{i=1}^{k}A_i\}}p_j = \sum\limits_{i=1}^{k}\sum_{\{j\mid s_j\in A_i\}}p_j\text{,}$$
and this last step is justified by the disjointedness of the $A_i$s, according to the text. But to do this, doesn't this assume that $\mathbb{P}$ is a probability measure to begin with, and thus creates a problem of circular logic?
 A: No, it doesn't use the fact that $\mathbb P$ is a probability. It uses the fact that, since $A_1,\cdots,A_k$ are pairwise disjoint, $s_j\in\bigcup_{i=1}^kA_i\Longleftrightarrow \exists!\, i_j(s_j\in A_{i_j})$.
In general, for finite sets $B_1,\cdots, B_k$, it holds
$$\sum_{i=1}^k\sum_{x\in B_i}g_x=\sum_{x\in\bigcup_{i=1}^kB_i}g_x\cdot\#\{i:x\in B_i\}$$
which needs not be equal $\sum_{x\in\bigcup_{i=1}^kB_i}g_x$, if the $B_i$-s are not pairwise disjoint.
Of course, pairwise disjointness brings that $\forall x\in\bigcup_{i=1}^kB_i,\,\#\{i:x\in B_i\}=1$
A: Not really circular logic: the last equality
$$ \sum\limits_{\{j\mid s_j \in \bigcup_{i=1}^{k}A_i\}}p_j = \sum\limits_{i=1}^{k}\sum_{\{j\mid s_j\in A_i\}}p_j\text{,}$$
is a statement about summing some real numbers that happen to satisfy certain conditions, and doesn't involve the prob measure $\mathbb P$. We can write this out in more detail:
$$\sum\limits_{\{j\mid s_j \in \bigcup_{i=1}^{k}A_i\}}p_j 
=\sum_jp_jI(s_j\in\bigcup_iA_i)\stackrel{(1)}=
\sum_jp_j \sum_i I(s_j\in A_i)\stackrel{(2)}=\sum_i\sum_jp_jI(s_j\in A_i)
$$
where the indicator $I(A)$ has value 1 if the statment $A$ is true, 0 otherwise. In step (1) we use the disjointness of the $A_i$'s. In step (2) we are interchanging the order of a double sum of a finite number of terms.
A: Main cause of your confusion is in my view the use of indexset $I=\left\{ 1,\cdots,n\right\} $
equipped with functions $I\rightarrow S$ (prescribed by $i\mapsto s_{i}$)
and $I\rightarrow\mathbb{R}$ (prescribed by $i\mapsto p_{i}$). 
Any set that has the same cardinality as $S$ can be used as indexset,
so we can do it with $S$ itself. This time equipped with the functions
prescribed by $s\mapsto s$ and $s\mapsto p_{s}$.
In the sequel I will handle the more general case where $S$ is countable
(in the sense that it is allowed to be finite) and non-empty.
For each $s\in S$ let $p_{s}$ be nonnegative in such a way that $\sum_{s\in S}p_{s}=1$.
For $A\subseteq S$ define: $\mathbb{P}\left(A\right)=\sum_{s\in A}p_{s}$.
Here an empty summation is identified with $0$ so: $\mathbb{P}\left(\varnothing\right)=0$.
It is evident that $\mathbb{P}\left(A\right)\geq0$ for each $A\subseteq S$
and that $\mathbb{P}\left(S\right)=\sum_{s\in S}p_{s}=1$. 
If $A_{1},A_{2},\dots$
denote disjoint subsets of $S$ then:
$$\mathbb{P}\left(\bigcup_{i=1}^{\infty}A_{i}\right)=\sum_{s\in\bigcup_{i=1}^{\infty}A_{i}}p_{s}=\sum_{i=1}^{\infty}\sum_{s\in A_{i}}p_{s}=\sum_{i=1}^{\infty}\mathbb{P}\left(A_{i}\right)$$
The second equality is legal and proved is now that $\mathbb{P}$ is a probability measure.
In the special case of a finite $S$ there will exist an integer $k$
such that $i>k\implies A_{i}=\varnothing$ so that $\bigcup_{i=1}^{\infty}A_{i}=\bigcup_{i=1}^{k}A_{i}$.
