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.

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Is this homework? – Generic Human Jun 23 '12 at 1:47
No, this is my original question. – Makoto Kato Jun 23 '12 at 3:08

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.

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+1 For punching through this horrible mess – Norbert Jun 23 '12 at 15:21
$(A\times B) \setminus (C\times D)=((A\setminus C)\times B)\ \,\cup\,((A\cap C)\times (B\setminus D))$. – vesszabo Aug 30 '13 at 14:41