Let X be a set. Let $\Delta$ be a semiring on X. Let $\mu$ be a function $\Delta \rightarrow \mathbb{R}$. We say $\mu$ is a finitely additive function on $\Delta$ if it satisfies the following condition (*).
(*) Let $A, A_1、...、A_m \in \Delta$. Suppose $A$ is a disjoint union of $A_1、...、A_m$. Then $\mu(A) = \mu(A_1) + ... + \mu(A_m)$
Let X and Y be sets. Let $\Delta$ and $\Gamma$ be semirings on X and Y respectively. Let $\mu$ and $\nu$ be finitely additive functions on X and Y respectively. Let $\Sigma$ be the set {$A \times B: A \in \Delta, B \in \Gamma$}.
Define a function $\lambda: \Sigma \rightarrow \mathbb{R}$ by $\lambda(A \times B) = \mu(A) \times \nu(B)$.
Then, is the following statement true?
$\Sigma$ is a semiring on $X \times Y$ and $\lambda$ is finitely additive on it.
EDIT I think the above result follows immediately from the following lemma whose proof I hope is correct.
Lemma Let $X$ be a set. Let $\Delta$ be a semiring on X. Let $I_1, ..., I_n$ be elements of $\Delta$. Then there exist $J_1, ..., J_m \in \Delta$ with the following properties.
(1) $J_1, ..., J_m$ are mutually disjoint.
(2) $I_1 \cup ... \cup I_n = J_1 \cup ... \cup J_m$
(3) Each $I_k$ is a union of a subset of {$J_1, ... ,J_m$}.
Proof: Let $A = I_1 \cup ... \cup I_n$. Let $S = \{1, 2, ... , n\}$. Let $T = \{i_1, ... , i_r\}$ be a non-empty subset of $S$. Let $\{j_1, ... , j_s\} = S - T$. Let $J_T = I_{i_1} \cap ... \cap I_{i_r} \cap (A - I_{j_1}) \cap ... \cap(A - I_{j_s})$. Let $P(S)$ be the power set of $S$. Then $I_1 \cup ... \cup I_n$ is the disjoint union of {$J_T; T \in P(S) - \{\emptyset\}$}. Each $I_k$ is a union of a subset of {$J_T; T \in P(S) - \{\emptyset\}$}. Since $\Delta$ is a semiring, each $J_T$ is a disjoint union of elements of $\Delta$. QED
EDIT This is an application of the above result.