Proving finite additivity for this semi-algebra (infinite coin flips) Background copied and pasted from another one of my questions:
Background: Consider flipping a coin $n$ times. Define the sample space as
$$
\Omega = \{(r_1,r_2,r_3,\dots); r_i = 0 \text{ or }1\}
$$
Define subsets of the sample space as
$$
A_{a_1a_2\dots a_n} = \{(r_1,r_2,\dots )\in \Omega; r_i =a_i \text{ for } 1\leq i \leq n\}
$$
where $r_i$ is $0$ if the $i$th coin flip is tails and $1$ if it is heads.
Define a set $\mathcal{J}$ by
$$
\mathcal{J} = \{ A_{a_1a_2\dots a_n}; n\in \mathbb{N}, a_1,a_2,\dots ,a_n \in \{ 0,1\}\} \cup \{\emptyset , \Omega\}
$$
Let $P(A_{a_1a_2\dots a_n}) = 1/2^n$ for each set $A_{a_1a_2\dots a_n}$
Problem: I want to show that the above $\mathcal{J}$ and $P$ satisfy the following property for finite collections $\{D_n\}$:
$$
P(\cup_n D_n)= \sum_n P(D_n) \text{ for } D_1,D_2,\dots \in \mathcal{J} \text{ disjoint with }\cup_nD_n \in \mathcal{J}
$$
Hint: The hint I am given is: For a finite collection $\{D_n\}\in\mathcal{J}$, there is a $k\in \mathbb{N}$ such that the results of only coins $1$ through $k$ are specified by any $D_n$. Partition $\Omega$ into the corresponding $2^k$ subsets.
I have submitted what I think a solution, yet I would be very happy if someone can provide me with a cleaner/shorter solution (I believe one exists).
Thanks.
Edit: Note: I changed my interpretation of $k$ from what I thought initially.
Edit: I have deleted my work since I think I found a solution and now simply want a nicer one. I think the cleaner look makes it more likely that I will get one.
 A: Yes, you are right in your belief, and the following answer is based on the hint. 
To simplify the answer, we shall use the following notations. For each natural $k$ let $\{0,1\}^k$ be the set of $0$-$1$ words of length $k$. For the convenience, as $\{0,1\}^0$ we denote the set consisting of empty word $\varnothing$. Put $\{0,1\}^\infty=\bigcup_{n=0}^\infty \{0,1\}^n$. Let $v,w\in \{0,1\}^\infty$  be words. We shall write $v\le w$, if the word $v$ is a prefix of the word $w$, that is if $w=vu$ for some (possibly, empty) word $u\in \{0,1\}^\infty$. In our notation, $\mathcal J=\{\varnothing\}\cup\bigcup_{v\in \{0,1\}^\infty} A_v$ (remark that $\Omega= A_\varnothing$).  
Let $\mathcal D$ be a finite subcollection of the family $\mathcal J\setminus\{\varnothing\} $. Then there exists a finite subset $V$ of $\{0,1\}^\infty$ such that  $\mathcal D=\{A_v:v\in V\}$. Since the collection $\mathcal D$ is finite, there exists a natural number $N$ such that $V\subset \bigcup_{n=0}^N \{0,1\}^n$. It is easy to check that $A_v=\bigcup \{A_w: w\in  \{0,1\}^N$ and $v\le w\}$ and $$P(A_v)=\sum \{P(A_w): w\in  \{0,1\}^N\mbox{ and }v\le w\}$$ for each $v\in V$. Put $W=\{w\in \{0,1\}^N:\exists v\in V: v\le w\}$. Put $A=\bigcup\mathcal D=\bigcup \{A_v:v\in V\}$. Then $A=\bigcup \{A_w: w\in W\}$. If $A\in\mathcal J$ then $A=A_u$ for some $u\in\{0,1\}^\infty$ and it is easy to check that $W=\{w\in \{0,1\}^N:u$ is a prefix of $w\}$. Then $$\sum_{D\in\mathcal D} P(D)=\sum_{v\in V} P(A_v)=\sum_{w\in W} P(A_w)=P(A_u)$$ (remark that the collection $\mathcal D $ should be disjoint in order to satisfy the second equality).
A: Let $\{D_n\}$ be a finite collection of $D_i \in \mathcal{J}$. Note that every element in $\mathcal{J}$ is of the form $A_{a_1a_2\dots a_m}$, so every element of the collection $\{D_n\}$ has some length (I'm using length as : the length of $A_{a_1a_2 \dots a_m}$ is $m$).
Let $k$ be the smallest length in $\{D_n\}$ (it is the smallest $m$). Denote the element with this length by $L$. Now, partition $\Omega$ into $2^k$ disjoint partitions, denoted $B_j, j\in \{1,2,\dots, 2^k) = J$. At least one of these is $L$, denote it $B_{j_L}$.
Consider 
$$\sum_{j=1}^{2^k}P(B_j) = P(B_{j_L})+ \sum_{j\in J\setminus j_L}P(B_j) =1$$
 If there exists any other $D_i \in \{D_n\}$ (besides $B_{j_L}$) such that $D_i = B_j, j\not = j_L$ denote them by $B_{j_{s}}, s\in S$, where $S$ is a set containing the indexes of all such elements.
To simplify the notation, let $S' = S\cup j_L$. That is, $S'$ is $S$ with the index for $j_L$ added. Let $q =|S'|$, the cardinality of $S'$ Now we have:
$$
\sum_{j=1}^{2^k}P(B_j) = \sum_{i\in S'}P(B_i)+ \sum_{j\in J\setminus S'}P(B_j) =1
$$
For any remaining $D_i \in \{D_n\}$ s.t. $D_i \not = B_i, i\in S'$, let $\{B_l\}$ be the set of all $B_j$ whose $k$ coin toss results match the first $k$ coin toss results of at least one $D_i \in \{D_n\}$ Let $Q$ denote the set of indices for the $B_l$'s; that is, let $\{B_l\} = \{B_q\}_{q \in Q}$
For every $B_q q\in Q$, Let $t_q$ be the sequence $a_1a_2\dots a_m$ of $D_i$, for whichever $D_i$ shares the same $k$ coin results with $B_q$. If $B_q$ shares the first $k$ results with more than one $D_i$, choose the longest $t_q$. Let $k_q$ be the length of each $t_q$ Let $\Omega_{k_q} = \Omega = \{(r_1,r_2,r_3,\dots r_{k_q}); r_i = 0 \text{ or }1\}$. Note that $ \forall q, B_q = \cup_{i\in \Omega_{k_q}} A_i$ Therefore,
$$
\sum_{j=1}^{2^k}P(B_j) = \sum_{i\in S'}P(B_i)+ \sum_{q\in Q}\sum_{i\in\Omega_{k_q}}P(A_i) \sum_{j\in J\setminus S''}P(B_j) = 1
$$
where $S'' = S'\cup Q$. We can write this as
$$
\sum_{i\in S'}P(B_i)+ \sum_{q\in Q}\sum_{i\in\Omega_{k_q}}P(A_i\setminus \{D_i\}) + \sum_{q\in Q}\sum_{i\in\Omega_{k_q}}P(A_i\setminus \{D_i\}^c) + \sum_{j\in J\setminus S''}P(B_j)=1
$$
rearranging further gives 
$$\sum_{i\in S'}P(B_i) + \sum_{q\in Q}\sum_{i\in\Omega_{k_q}}P(A_i\setminus \{D_i\}^c) =1- \sum_{j\in J\setminus S''}P(B_j) - + \sum_{q\in Q}\sum_{i\in\Omega_{k_q}}P(A_i\setminus \{D_i\})$$
The RHS which is $P(\cup D_i), D_i \in \{D_n\}$, because it is one minus the probability of the complement
A: Here is my solution (re-posted from my blog).
PROBLEM STATEMENT:
Suppose that we wish to model the flipping of a countably infinite number of coins.  Let the sample space $\Omega$  be $\{(r_{1},r_{2},\ldots) \vert r_{i} = 0$ or $1\}$.  
For each $n \in \mathbb{N}$ and each $a_{1},\ldots,a_{n} \in \{0,1\}$, let $A_{a_{1}\ldots a_{n}} \subseteq \Omega$ be $\{(r_{1},r_{2},\ldots) \in \Omega \vert r_{i} = a_{i}$ for $i \leq i \leq n\}$.  It is straightforward to show that the set of all such $A_{a_{1}\ldots a_{n}}$ along with $\{\emptyset,\Omega\}$ forms a semialgebra.  Call this set $\mathbb{I}$.  Define a function $P$ on this set with $P(A_{a_{1}\ldots a_{n}})=\frac{1}{2^n}$, $P(\emptyset)=0$, and $P(\Omega)=1$.  
The exercise is to show that $P$ is "finitely additive" on $\mathbb{I}$ in the following sense: whenever $A=\cup_{i=1}^{n}A_{i}$ with $A,A_{1},\ldots ,A_{n}$ disjoint elements of $\mathbb{I}$, then $P(A)=\Sigma_{i=1}^{n} P(A_{i})$.  (This is a step towards applying Carathéodory's extension theorem, which allows us to conclude that $P$ can be extended to a probability measure defined on a $\sigma$-algebra $\mathbb{F}$ with $\mathbb{I} \subseteq \mathbb{F}$.)
SOLUTION:
Say that the "length" of an element $A_{a_{1}\ldots a_{n}}$ of $\mathbb{I}$ is $n$.  (Also define the length of $\emptyset$ to be 0 and the length of $\Omega$ to be $\infty$.)  For any $k>n$, $A_{a_{1}\ldots a_{n}}$ is the disjoint union of all sets of the form $A_{a_{1}\ldots a_{n} a_{n+1} \ldots a_{k}}$, where $a_{n+1} \ldots a_{k}$ ranges over all possible assignments of $0$ and $1$ to those values.  Notice that there are $2^{(k-n)}$ such sets, each of which has $P(A_{a_{1}\ldots a_{n} a_{n+1} \ldots a_{k}})=\frac{1}{2^{k}}$.  Label these sets (in whatever order you like) as $B_j$, $j \in \{1,2,\ldots,2^{(k-n)}\}$.  It follows, then, that $P(A_{a_{1}\ldots a_{n}} )=\frac{1}{2^n}=\Sigma_{j=1}^{2^{(k-n)}}P(B_{j})$, since $\Sigma_{j=1}^{2^{(k-n)}}P(B_{j})=2^{k-n}\frac{1}{2^k}=\frac{1}{2^n}$.  To put this observation informally, the probabilities work out "correctly" when you decompose a length $n$ element into a disjoint union of length $k$ elements, where $k>n$.
Suppose that A is a non-empty set such that $A=\cup_{i=1}^{n}A_{i}$ with $A,A_{1},\ldots ,A_{n} \in \mathbb{I}$.  Let $k$ be the maximum length of $A,A_{1},\ldots ,A_{n}$.  Decompose each $A,A_{1},\ldots ,A_{n}$ into a disjoint union of elements of $\mathbb{I}$, all of which have length $k$.  Let the length $k$ sets associated with $A_{i}$ be called $B_{ij}$.  Say that the number of such sets involved in the decomposition of $A_{i}$ is $n_{i}$.
We have already shown above that $P(A_{i})=\Sigma_{j=1}^{n_{i}}P(B_{ij})$.  Now, if we decompose $A$ into a disjoint union of length $k$ elements of $\mathbb{I}$, these length $k$ elements must be precisely $\{B_{ij}\}$.  Thus, $P(A)=\Sigma_{i,j}P(B_{ij})$.  If we group the terms of this sum appropriately and use the identity $P(A_{i})=\Sigma_{j=1}^{n_{i}}P(B_{ij})$, we obtain the identity that we wanted: $P(A)=\Sigma_{i=1}^{n} P(A_{i})$.
The argument above assumed that $A$ was non-empty.  The case of $A=\emptyset$ is trivial, since the only disjoint union of element of $\mathbb{I}$ that equals $A=\emptyset$ is the union of $\emptyset$ and nothing else, in which case finite additivity also holds.  
