(a) Definition of the σ-field generated by a class $A$ of subsets of $\Omega$:

The σ-field (or σ-algebra) generated by a class $A$ of subsets of a set $\Omega$ is the smallest σ-field that contains $A$. Denoted as $\sigma(A)$, it is the intersection of all σ-fields that contain $A$. Formally,

$\sigma(A) = \bigcap \{ \mathcal{F} \mid A \subseteq \mathcal{F} \text{ and } \mathcal{F} \text{ is a σ-field on } \Omega \}.$

(b) Proof that if $A \subset B \subset \sigma(A)$, then $\sigma(A) = \sigma(B)$:

Let $\sigma(A)$ and $\sigma(B)$ denote the σ-fields generated by $A$ and $B$, respectively.

1. Since $A \subseteq B$:
   By the definition of the generated σ-field, $\sigma(A) \subseteq \sigma(B)$, because $\sigma(B)$ is the smallest σ-field containing $B$, and $B$ contains $A$.

2. Since $B \subseteq \sigma(A)$:
   Every σ-field containing $A$ must also contain $B$ because $A \subseteq B$. Therefore, $\sigma(A)$ is also the smallest σ-field containing $B$, implying $\sigma(B) \subseteq \sigma(A)$.

3. Conclusion:
   From $\sigma(A) \subseteq \sigma(B)$ and $\sigma(B) \subseteq \sigma(A)$, it follows that $\sigma(A) = \sigma(B)$.

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(c) Show that the intersection of any number of σ-fields is a σ-field and redefine $\sigma(A)$:

Step 1: Show the intersection of σ-fields is a σ-field

Let $\{ \mathcal{F}_\alpha \}_{\alpha \in I}$ be a collection of σ-fields on $\Omega$, where $I$ is an index set. Define

$\mathcal{F} = \bigcap_{\alpha \in I} \mathcal{F}_\alpha.$

We show that $\mathcal{F}$ is a σ-field:

1. Closure under $\Omega$:
   Since each $\mathcal{F}_\alpha$ is a σ-field, $\Omega \in \mathcal{F}_\alpha$ for all $\alpha$. Hence, $\Omega \in \mathcal{F}$.

2. Closure under complementation:
   If $A \in \mathcal{F}$, then $A \in \mathcal{F}_\alpha$ for all $\alpha$. Since each $\mathcal{F}_\alpha$ is a σ-field, $A^c \in \mathcal{F}_\alpha$ for all $\alpha$. Therefore, $A^c \in \mathcal{F}$.

3. Closure under countable unions:
   If $A_1, A_2, \dots \in \mathcal{F}$, then $A_n \in \mathcal{F}_\alpha$ for all $\alpha$ and all $n$. Since each $\mathcal{F}_\alpha$ is a σ-field, $\bigcup_{n=1}^\infty A_n \in \mathcal{F}_\alpha$ for all $\alpha$. Hence, $\bigcup_{n=1}^\infty A_n \in \mathcal{F}$.

Thus, $\mathcal{F}$ is a σ-field.

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Step 2: Redefine $\sigma(A)$

Using the above result, the σ-field generated by $A$ can be redefined as:

$\sigma(A) = \bigcap \{ \mathcal{F} \mid A \subseteq \mathcal{F} \text{ and } \mathcal{F} \text{ is a σ-field on } \Omega \}.$

This emphasizes that $\sigma(A)$ is the smallest σ-field containing $A$.

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Questions to Expand Understanding:

1. What are some examples of sets $A$ and their corresponding $\sigma(A)$ in probability theory?
2. Why is the intersection of σ-fields used to define $\sigma(A)$?
3. Can you prove that $\sigma(A)$ is unique for a given $A$?
4. How does the concept of a σ-field apply to measurable functions?
5. What is the relationship between $\sigma(A)$ and the Borel σ-field in $\mathbb{R}$?

Tip: When working with σ-fields, always verify the three properties (containment of $\Omega$, closure under complements, and countable unions) to ensure correctness!