Finitely additive function on a product of semirings of sets

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.

Solutions Collecting From Web of "Finitely additive function on a product of semirings of sets"

That $\Sigma$ is a semiring on $X\times Y$ follows directly from the definition. For difference, you just have to notice that $(A\times B) \setminus (C\times D)=(A\setminus C)\times B\ \,\cup\,\ C\times (B\setminus D)$.

Now for finite additivity of $\lambda$. Suppose we have the disjoint union:
$$A\times B=\bigcup_{i=1}^n A_i\times B_i$$

We will prove additivity of $\lambda$ by slicing $A_i\times B_i$ along a sufficiently fine partition of $A$, so that we can then apply finite additivity of $\mu$ along each axis separately.

Define $F$ the (finite) set of finite intersections of $A_i$, including the $A_i$ themselves. Let $S$ be the set of minimal non-empty elements of $F$ with respect to inclusion. Then for any point $a$ in $A$, the intersection of all sets of $F$ containing $a$ belongs to $S$, so that the union of $S$ is $A$ and because all elements of $S$ are pairwise disjoint, $S$ forms a partition of $A$.
For $x\in S$:
$$\begin{align}
x\times B=&\bigcup_{i=1}^n (A_i\cap x)\times B_i\\
=&\bigcup_{A_i\cap x\ne\emptyset} x\times B_i
\end{align}$$
proving that for all $x\in S$, $\bigcup_{A_i\cap x\ne\emptyset} B_i=B$.

For all $i$, $A’_i=\displaystyle\bigcup_{\substack{x\in S\\A_i\cap x\ne\emptyset}}x$ is both a superset of $A_i$ because $S$ is a partition of $A$, and a subset of $A_i$ because $x\cap A_i\ne\emptyset$ implies $x\cap A_i=x$ and therefore $x\subseteq A_i$, so that $A’_i=A_i$.

We then have
$$\begin{align}
\lambda(A\times B)=&\mu(A)\times\mu(B)\\
=&\mu\left(\textstyle\bigcup_{x\in S} x\right)\times\mu(B)\\
=&\sum_{x\in S}\mu(x)\times\mu(B)\\
=&\sum_{x\in S}\mu(x)\times\sum_{A_i\cap x\ne\emptyset}\mu(B_i)\\
=&\sum_{i=1}^n\sum_{\substack{x\in S\\A_i\cap x\ne\emptyset}}\mu(x)\times\mu(B_i)\\
=&\sum_{i=1}^n\mu\Big(\bigcup_{\substack{x\in S\\A_i\cap x\ne\emptyset}}x\Big)\times\mu(B_i)\\
=&\sum_{i=1}^n\mu(A_i)\times\mu(B_i)\\
=&\sum_{i=1}^n\lambda(A_i\times B_i)\\
\end{align}$$
so we have finite additivity.