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First of all, I would like to say that I do not see any good reason not to use Stone spaces in order to describe the coproduct of Boolean algebras. The product of Stone spaces is the most natural way to represent free products of Boolean algebras. However, besides Stone spaces, there are a couple other ways to represent the free product of Boolean algebras.

First of all, if you represent Boolean algebras in terms of algebras of sets, then the free product is easy to describe. Assume that $(X_{i},\mathcal{A}_{i})$ is an algebra of sets for $i\in I$. Let $X=\prod_{i\in I}X_{i}$ and let $\pi_{i}:X\rightarrow X_{i}$ be the projection mappings. Let $\mathcal{A}$ be the algebra of sets over $X$ generated by sets of the form $\pi_{i}^{-1}[R]$ for $i\in I$. Then $\mathcal{A}$ is the free product of the Boolean algebras $\mathcal{A}_{i}$. Furthermore, every element in $\mathcal{A}$ can be written as a finite disjoint union of sets of the form $\pi_{i_{1}}^{-1}[R_{1}]\cap...\cap\pi_{i_{n}}^{-1}[R_{n}]$ for $R_{1}\in\mathcal{A}_{1},...,R_{n}\in\mathcal{A}_{n}$ (This is because the sets of the form $\pi_{i_{1}}^{-1}[R_{1}]\cap...\cap\pi_{i_{n}}^{-1}[R_{n}]$ form a semialgebra of sets).

Also, the free product of two Boolean algebras is the Boolean power of those two Boolean algebras (i.e. $A\ast B\simeq A^{B}$ where $A^{B}$ denotes the Boolean power). The Boolean power $A^{B}$ is the set of all continuous functions $f:S(B)\rightarrow A$ where $A$ is given the discrete topology and $S(B)$ is the Stone space of $B$. The Boolean power $A^{B}$ can also be described as the direct limit $^{\lim}_{\longrightarrow}A^{|p|}$ where $p$ ranges over the set $\mathbb{P}_{\omega}(B)$ of finite partitions of $B$ ordered by reverse refinement. In other words, $p\leq q$ if for each $a\in q$ there is a $b\in p$ with $a\leq b$. The transitional mappings $A^{|p|}\rightarrow A^{|q|}$ are the natural surjections whenever $p\leq q$. See the book A Course in Universal Algebra by Stanley Burris and H.P. Sankappanaver for more details on the Boolean power.

Now, in order to effortlessly put a finitely additive measure on the product of Stone spaces, we will need to use Stone duality and a couple results from measure theory. In essense we use the product of countably additive measures on $\sigma$-algebras to give us the product of finitely additive measures on Boolean algebras.

Assume that for $i\in I$, $B_{i}$ is a Boolean algebra and $\mu_{i}$ is a finitely additive measure on $B_{i}$. Then by the Caratheodory Extension Theorem, there is a Baire measure $\nu_{i}$ on the Stone space $S(B_{i})$ such that $\nu_{i}(\{\mathcal{U}\in S(B_{i})|a\in\mathcal{U}\})=\mu_{i}(a)$ for $a\in B_{i}$. Let $\nu$ be the product measure of the measures $\nu_{i}$ on the product space $\prod_{i\in I}S(B_{i})$. Then we get a measure on the free product $\ast_{i\in I}B_{i}$ simply by restricting $\nu$ to the clopen sets of $\prod_{i\in I}S(B_{i})$.

Also, the free product of Boolean algebras is completely described by the following property. Assume that $B$ is a Boolean algebra and $B_{i}\subseteq B$ is a subalgebra for $i\in I$. The system of subalgebras $(B_{i})_{i\in I}$ is said to be independent if whenever $i_{1},....,i_{n}\in I$ are distinct elements and $b_{i}\in B_{i},b_{i}>0$ for $i\in\{i_{1},...,i_{n}\}$, then $b_{i_{1}}\wedge...\wedge b_{i_{n}}>0$. The Boolean algebra $B$ is the free product of the Boolean algebras $(B_{i})_{i\in I}$ if and only if the Boolean algebras $B_{i}$ are independent and $B$ is generated by $\bigcup_{i\in I}B_{i}$.
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First of all, I would like to say that I do not see any good reason not to use Stone spaces in order to describe the coproduct of Boolean algebras. The product of Stone spaces is the most natural way to represent free products of Boolean algebras. However, besides Stone spaces, there are a couple other ways to represent the free product of Boolean algebras.

First of all, if you represent Boolean algebras in terms of algebras of sets, then the free product is easy to describe. Assume that $(X_{i},\mathcal{A}_{i})$ is an algebra of sets for $i\in I$. Let $X=\prod_{i\in I}X_{i}$ and let $\pi_{i}:X\rightarrow X_{i}$ be the projection mappings. Let $\mathcal{A}$ be the algebra of sets over $X$ generated by sets of the form $\pi_{i}^{-1}[R]$ for $i\in I$. Then $\mathcal{A}$ is the free product of the Boolean algebras $\mathcal{A}_{i}$. Furthermore, every element in $\mathcal{A}$ can be written as a finite disjoint union of sets of the form $\pi_{i_{1}}^{-1}[R_{1}]\cap...\cap\pi_{i_{n}}^{-1}[R_{n}]$ for $R_{1}\in\mathcal{A}_{1},...,R_{n}\in\mathcal{A}_{n}$ (This is because the sets of the form $\pi_{i_{1}}^{-1}[R_{1}]\cap...\cap\pi_{i_{n}}^{-1}[R_{n}]$ form a semialgebra of sets).

Also, the free product of two Boolean algebras is the Boolean power of those two Boolean algebras (i.e. $A\ast B\simeq A^{B}$ where $A^{B}$ denotes the Boolean power). In essense, the The Boolean power $A^{B}$ is the set of all continuous functions $f:S(B)\rightarrow A$ where $A$ is given the discrete topology and $S(B)$ is the Stone space of $B$. The Boolean power $A^{B}$ can also be described as the direct limit $^{\lim}_{\longrightarrow}A^{|p|}$ where $p$ ranges over the set $\mathbb{P}_{\omega}(B)$ of finite partitions of $B$ ordered by reverse refinement. In other words, $p\leq q$ if for each $a\in q$ there is a $b\in p$ with $a\leq b$. The transitional mappings $A^{|p|}\rightarrow A^{|q|}$ are the natural surjections whenever $p\leq q$. See the book A Course in Universal Algebra by Stanley Burris and H.P. Sankappanaver for more details on the Boolean power.

Now, in order to effortlessly put a finitely additive measure on the product of Stone spaces, we will need to use Stone duality and a couple results from measure theory. In essense we use the product of countably additive measures on $\sigma$-algebras to give us the product of finitely additive measures on Boolean algebras.

Assume that for $i\in I$, $B_{i}$ is a Boolean algebra and $\mu_{i}$ is a finitely additive measure on $B_{i}$. Then by the Caratheodory Extension Theorem, there is a Baire measure $\nu_{i}$ on the Stone space $S(B_{i})$ such that $\nu_{i}(\{\mathcal{U}\in S(B_{i})|a\in\mathcal{U}\})=\mu_{i}(a)$ for $a\in B_{i}$. Let $\nu$ be the product measure of the measures $(\nu_{i}){i\in I}$\nu_{i}$ on the product space $\prod{i\in \prod_{i\in I}S(B_{i})$. Then we get a measure on the free product $\ast_{i\in I}B_{i}$` I}B_{i}$ simply by restricting $\nu$ to the clopen sets of $\prod_{i\in I}S(B_{i})$.
show/hide this revision's text 1

First of all, I would like to say that I do not see any good reason not to use Stone spaces in order to describe the coproduct of Boolean algebras. The product of Stone spaces is the most natural way to represent free products of Boolean algebras. However, besides Stone spaces, there are a couple other ways to represent the free product of Boolean algebras.

First of all, if you represent Boolean algebras in terms of algebras of sets, then the free product is easy to describe. Assume that $(X_{i},\mathcal{A}_{i})$ is an algebra of sets for $i\in I$. Let $X=\prod_{i\in I}X_{i}$ and let $\pi_{i}:X\rightarrow X_{i}$ be the projection mappings. Let $\mathcal{A}$ be the algebra of sets over $X$ generated by sets of the form $\pi_{i}^{-1}[R]$ for $i\in I$. Then $\mathcal{A}$ is the free product of the Boolean algebras $\mathcal{A}_{i}$. Furthermore, every element in $\mathcal{A}$ can be written as a finite disjoint union of sets of the form $\pi_{i_{1}}^{-1}[R_{1}]\cap...\cap\pi_{i_{n}}^{-1}[R_{n}]$ for $R_{1}\in\mathcal{A}_{1},...,R_{n}\in\mathcal{A}_{n}$ (This is because the sets of the form $\pi_{i_{1}}^{-1}[R_{1}]\cap...\cap\pi_{i_{n}}^{-1}[R_{n}]$ form a semialgebra of sets).

Also, the free product of two Boolean algebras is the Boolean power of those two Boolean algebras (i.e. $A\ast B\simeq A^{B}$ where $A^{B}$ denotes the Boolean power). In essense, the Boolean power $A^{B}$ is the set of all continuous functions $f:S(B)\rightarrow A$ where $A$ is given the discrete topology and $S(B)$ is the Stone space of $B$. The Boolean power $A^{B}$ can also be described as the direct limit $^{\lim}_{\longrightarrow}A^{|p|}$ where $p$ ranges over the set $\mathbb{P}_{\omega}(B)$ of finite partitions of $B$ ordered by reverse refinement. In other words, $p\leq q$ if for each $a\in q$ there is a $b\in p$ with $a\leq b$. The transitional mappings $A^{|p|}\rightarrow A^{|q|}$ are the natural surjections whenever $p\leq q$. See the book A Course in Universal Algebra by Stanley Burris and H.P. Sankappanaver for more details on the Boolean power.

Now, in order to effortlessly put a finitely additive measure on the product of Stone spaces, we will need to use Stone duality and a couple results from measure theory. In essense we use the product of countably additive measures on $\sigma$-algebras to give us the product of finitely additive measures on Boolean algebras.

Assume that for $i\in I$, $B_{i}$ is a Boolean algebra and $\mu_{i}$ is a finitely additive measure on $B_{i}$. Then by the Caratheodory Extension Theorem, there is a Baire measure $\nu_{i}$ on the Stone space $S(B_{i})$ such that $\nu_{i}(\{\mathcal{U}\in S(B_{i})|a\in\mathcal{U}\})=\mu_{i}(a)$ for $a\in B_{i}$. Let $\nu$ be the product measure of$(\nu_{i}){i\in I}$on the product space$\prod{i\in I}S(B_{i})$. Then we get a measure on the free product$\ast_{i\in I}B_{i}$` simply by restricting $\nu$ to the clopen sets.