Finitely additive measure on cartesian product Let $M_{1}$ and $M_{2}$ be two basic sets and $S_{1}, S_{2}$ be the semi-rings in $M_{1}$ and $M_{2}$, respectively. Let $\mu_{1}$ and $\mu_{2}$ be finitely additive measures on $S_{1}$ and $S_{2}$, respectively. Let us define a function $\mu$ on the semi-ring $S_{1} \times S_{2}$ as follows: for $A=B \times C$ with $B \in S_{1}$ and $C \in S_{2}$ we set
$$
\mu(A)=\mu_{1}(B) \mu_{2}(C)
$$
Now the author says that $\mu$ is a finitely additive measure on $S_{1} \times S_{2}$ but I cannot see why. I need to show that if
$$
A:= \bigsqcup_k^n A_k \Rightarrow \mu(A)=\sum_k^n \mu(A_k)
$$
But I have trouble verifying it. Maybe someone has got a hint.
 A: Lemma:  Let $S$ be a set system in $M$ consisting of $N$ subsets. Then there is a
set system $T$ of at most $2^N-1$ disjoint subsets of $M$ such that $ρ(S)$ consists of the
elements of $T$ and all their possible unions where $ρ(S)$ is the intersection of all
rings containing $S$.

Consider the rings $R_i=\rho(S_i)$ and extend measure $\mu_i$ to $R_i$ for $i=1,2.$
We need to prove that if $A\in S_1 \times S_2$ is a disjoint union of a finite number of sets $A_1, A_2,...,A_N \in S_1 \times S_2, $ then
$$
\mu(A)=\sum_{i-1}^{N} \mu(A_{i})
$$
Let $A_{i}=B_{i} \times C_{i}$ where $B_{i} \in S_{1}$ and $C_{1} \in S_{2}$. Clearly, "projections" of $A$ into $M_1$ and $M_2$ are $B=\bigcup_{i}B_i$ and $C=\bigcup_{k}C_k,$ respectively. Since $A$ has a form of a direct product, we see that $A=B \times C.$
By our Lemma, there is a finite number of disjoint sets in the ring $\rho(\{B_i\})$ such that each $B_i$ is the union of some of them. Denote those sets by $\{b_j\}.$ We can assume that
$$
B=\bigsqcup_j b_j
$$
(since we can throw away those $b_j$ that are not subsets of $B$). Similarly we find such sets for the family $\{C_i\}$ and denote them by $\{c_k\}$ so that
$$
C=\bigsqcup_k c_k.
$$
By construction, $b_j \in R_1$ and $c_k \in R_2.$ Then we obtain
\begin{equation}
    \mu(A)=\mu_1(B)\mu_2(C)=\sum_j \mu_1(b_j)\sum_k \mu_2(c_k)=\sum_{j,k}\mu_1(b_j)\mu_2(c_k).
    \label{eqn_001}
\end{equation}
By the choice of $\{b_j\}$ and $\{c_k\}$ we have for any $i$
$$
B=\bigsqcup_{b_j\subset B_i} b_j \hspace{2mm} \text{and} \hspace{2mm} C=\bigsqcup_{c_k\subset C_i} c_k,
$$
whence
\begin{equation}
    \mu(A_i)=\mu_1(B_i)\mu_2(C_i)=\sum_{b_j\subset B_i} \mu_1(b_j)\sum_{c_k\subset C_i} \mu_2(c_k)=\sum_{j,k}\mu_1(b_j)\mu_2(c_k).
    \label{eqn_002}
\end{equation}
Finally, by comparing the above lines, we obtain
$$
\sum_i \mu(A_i)=\sum_i \sum_{b_j\subset B_i, c_k\subset C_i } \mu_1(b_j)\mu_2(c_k) =\sum_{j,k} \mu_1(b_j)\mu_2(c_k) = \mu(A). \quad \blacksquare
$$
