Hey I want to check my solutions for this problem:
Consider the sample space $\Omega = \{0, 1\}^N$ of an infinitely repeated coin toss. Let $Π_n: \Omega \rightarrow \{0, 1\}^n$ be the coordinate projection defined by $Π_n(ω) = (ω_1, ..., ω_n)$.
(a) Show that the system of cylinder sets $A := \{Π_n^{-1}(A_n) | n \in \mathbb{N}, A_n \subseteq \{0, 1\}^n\}$ forms an algebra over $\Omega$.
(b) For a fair coin toss, define $P(Π_n^{-1}(A_n)) := |A_n|/2^n$. Demonstrate that $P$ is well-defined on $A$ and forms a normalized, additive set function.
(c) Prove that $P$ defines a probability measure on $A$ and can be extended to a measure on a σ-algebra that includes $A$. Hint: Use arguments analogous to the construction of the Lebesgue-Stieltjes measure.
So my solutions for a) To show that $A$ is an algebra we must prove:
$\Omega \in A$
$A_1, A_2 \in A \Rightarrow A_1 \cup A_2 \in A$
$A_1 \in \Rightarrow A_1^c \in A $
So we have
$\Omega \in A$ since $\Omega= Π_n^{-1}(A_n)$ for $A_n=\{0,1 \}^n$
Let $B_n \subseteq \{0,1\}^n$ and $C_m \subseteq \{0,1\}^m$ and $B=Π_n^{-1}(B_n)$ and $C=Π_n^{-1}(C_m)$ than we have $B \cup C=Π_n^{-1}(C_m) \cup Π_n^{-1}(C_m)=Π_n^{-1}(B_n x C_m)= \{0,1\}^{n+m}$
Let $A_n \subseteq \{0,1 \}^n$ with $A=Π_n^{-1}(A_n)$ $A_n^c \subseteq \{0,1\}^n \backslash A_n$ $A^c=Π_n^{-1}(\{0,1\}^n\backslash A_n) \subseteq \{0,1\}^n$
(b) We need to show that the function $P$ is well-defined on $A$ and forms a normalized, additive set function.
To prove well-definedness, we must show that for any $A_n \subseteq \{0,1\}^n$, the value of $P(Π_n^{-1}(A_n)) := |A_n|/2^n$ does not depend on the specific choice of $A_n$.
Consider two sets $A_n$ and $B_n$ such that $A_n \subseteq {0,1}^n$ and $B_n \subseteq \{0,1\}^n$. If $A_n = B_n$, then clearly $|A_n| = |B_n|$ and thus $P(Π_n^{-1}(A_n)) = P(Π_n^{-1}(B_n))$.
Now, suppose $A_n$ and $B_n$ are distinct sets. Since $A_n$ and $B_n$ have the same cardinality (i.e., the same number of elements), we have $|A_n| = |B_n|$. Consequently, $P(Π_n^{-1}(A_n)) = |A_n|/2^n = |B_n|/2^n = P(Π_n^{-1}(B_n))$. Hence, $P$ is well-defined on $A$.
To show that $P$ is a normalized, additive set function, we need to demonstrate two properties:
(i) $P(\Omega) = 1$: Since $\Omega = Π_n^{-1}({0,1}^n)$, we have $P(\Omega) = |\{0,1\}^n|/2^n = 2^n/2^n = 1$.
(ii) Additivity: Let $A_n$ and $B_n$ be two disjoint sets in $\{0,1\}^n$. We want to show that $P(Π_n^{-1}(A_n \cup B_n)) = P(Π_n^{-1}(A_n)) + P(Π_n^{-1}(B_n))$.
Using the definition of $P$, we have: $P(Π_n^{-1}(A_n \cup B_n)) = |A_n \cup B_n|/2^n$
Since $A_n$ and $B_n$ are disjoint, $|A_n \cup B_n| = |A_n| + |B_n|$. Hence, $P(Π_n^{-1}(A_n \cup B_n)) = (|A_n| + |B_n|)/2^n = |A_n|/2^n + |B_n|/2^n = P(Π_n^{-1}(A_n)) + P(Π_n^{-1}(B_n))$.
Therefore, $P$ is a well-defined, normalized, and additive set function on $A$.
(c) To prove that $P$ defines a probability measure on $A$ and can be extended to a measure on a σ-algebra that includes $A$, we need to show two things:
(i) $P$ is non-negative: Since $P$ is defined as the ratio of cardinalities, it is evident that $P(A) \geq 0$ for any $A \in A$.
(ii) $P$ is σ-additive: Let $A_1, A_2, A_3, \ldots$ be a countable collection of disjoint sets in $A$. We want to show that $P\left(\bigcup_{i=1}^{\infty} A_i\right) = \sum_{i=1}^{\infty} P(A_i)$.
Since the sets $A_i$ are disjoint, their corresponding $A_n$'s are also disjoint. We have: $P\left(\bigcup_{i=1}^{\infty} A_i\right) = P\left(Π_n^{-1}\left(\bigcup_{i=1}^{\infty} A_{n,i}\right)\right) = \left|\bigcup_{i=1}^{\infty} A_{n,i}\right|/2^n$
Similarly, $P(A_i) = \left|A_{n,i}\right|/2^n$ for each $i$.
Since the $A_{n,i}$'s are disjoint, we have $\left|\bigcup_{i=1}^{\infty} A_{n,i}\right| = \sum_{i=1}^{\infty} \left|A_{n,i}\right|$ by the countable additivity of cardinalities.
Hence, $P\left(\bigcup_{i=1}^{\infty} A_i\right) = \sum_{i=1}^{\infty} P(A_i)$, proving that $P$ is σ-additive.
Thus, $P$ satisfies the properties of a probability measure on $A$. It can be extended to a measure on a σ-algebra that includes $A$, making $P$ a probability measure over the extended space.
Have I done any mistakes? Thank you so much for any help