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If \(\mathcal{M}\) is the \(\sigma\)-algebra generated by \(\mathcal{\varepsilon}\), then \(\mathcal{M}\) is the union of the \(\sigma\)-algebras generated by \(\mathcal{F}\) as \(\mathcal{F}\) ranges over all countable subsets of \(\mathcal{E}\). (Hint: Show that the latter object is a \(\sigma\)-algebra.)

Short Answer

Expert verified
\( \mathcal{M} \) is the union of \( \sigma\)-algebras generated by countable subsets of \( \mathcal{E} \).

Step by step solution

01

Understand the Problem Statement

We are given that \( \mathcal{M} \) is the \( \sigma\)-algebra generated by a family \( \mathcal{E} \). We need to show that \( \mathcal{M} \) is the union of all \( \sigma\)-algebras generated by countable subsets of \( \mathcal{E} \).
02

Identify Properties of \( \sigma\)-Algebras

A \( \sigma\)-algebra \( \mathcal{A} \) over a set \( X \) must be closed under countable unions, countable intersections, and complementation, containing the empty set. The \( \sigma\)-algebra generated by \( \mathcal{E} \), denoted as \( \sigma(\mathcal{E}) \), is the smallest \( \sigma\)-algebra containing \( \mathcal{E} \).
03

Construct \( \sigma \)-Algebra from Countable Subsets

Consider all possible countable subsets \( \mathcal{F} \) of \( \mathcal{E} \). For each countable subset \( \mathcal{F} \), generate the \( \sigma\)-algebra \( \sigma(\mathcal{F}) \). Each of these \( \sigma\)-algebras is closed under the formation rules of a \( \sigma\)-algebra.
04

Form the Union of \( \sigma \)-Algebras

Form the union of all \( \sigma(\mathcal{F}) \) where \( \mathcal{F} \) is a countable subset of \( \mathcal{E} \). Denote this union as \( \bigcup_{\mathcal{F} \subseteq \mathcal{E}, \mathcal{F} \text{ countable}} \sigma(\mathcal{F}) \).
05

Verify \( \mathcal{M} \) as a \( \sigma \)-Algebra

Show that \( \bigcup_{\mathcal{F} \subseteq \mathcal{E}, \mathcal{F} \text{ countable}} \sigma(\mathcal{F}) \) is indeed a \( \sigma\)-algebra. Since \( \mathcal{F} \) is countable, \( \sigma(\mathcal{F}) \) will be closed under the rules for formations of a \( \sigma\)-algebra. Consequently, unioning all \( \sigma(\mathcal{F}) \) will not break these properties and thus it forms a \( \sigma\)-algebra.
06

Conclude and Simplify the Required Equation

Conclude that \( \mathcal{M} = \bigcup_{\mathcal{F} \subseteq \mathcal{E}, \mathcal{F} \text{ countable}} \sigma(\mathcal{F}) \). Therefore, \( \mathcal{M} \) can be viewed as the union of \( \sigma\)-algebras generated from countable subsets of \( \mathcal{E} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

generated σ-algebra
In probability and measure theory, a \( \sigma \)-algebra is a mathematical structure that helps in understanding collections of subsets. These \( \sigma \)-algebras are essential for defining measures, which in turn help in formulating probability. But what is a 'generated \( \sigma \)-algebra'?

A generated \( \sigma \)-algebra, denoted as \( \sigma(\mathcal{E}) \), is the smallest \( \sigma \)-algebra containing a given collection of subsets \( \mathcal{E} \). This means any other \( \sigma \)-algebra containing \( \mathcal{E} \) must also contain this generated \( \sigma \)-algebra.

To generate this smallest \( \sigma \)-algebra, we must ensure it is closed under certain operations, such as:
  • Countable unions: Combining infinitely many subsets from the collection.
  • Countable intersections: Finding common elements across an infinite number of subsets.
  • Taking complements: Finding what is not included in each subset.
Thus, a generated \( \sigma \)-algebra organizes and stretches a collection \( \mathcal{E} \) into a full array of subsets, thereby providing structure and rules for measure-related operations.
countable subsets
Many times in mathematics, dealing with infinite collections can be challenging. Yet, countable subsets make this easier. A countable set refers to a set that can be mapped to natural numbers neatly.

Countable subsets are essential in constructing \( \sigma \)-algebras because, often, infinite operations like unions and intersections are performed over them. This is particularly important when discussing the union of \( \sigma \)-algebras formed by countable subsets of a set \( \mathcal{E} \).
  • If a set is countable, like natural numbers or even a subset of real numbers, then it's manageable via standard mathematical tools.
  • Even if a whole set is uncountable, such as all real numbers, breaking it down into countable subsets allows us to work with parts that can be easier to handle.
Understanding countable subsets helps in the efficient construction and verification of structures like \( \sigma \)-algebras, ensuring these math systems hold up under various operations.
properties of σ-algebra
A \( \sigma \)-algebra is indispensable in fields like measure theory, which supports probability and integration theories. Its defining properties ensure reliability and consistency in mathematical operations:
  • Closed under countable unions: If you take an infinite union of sets in a \( \sigma \)-algebra, the result is also in the same \( \sigma \)-algebra.
  • Closed under countable intersections: Any infinite intersection of sets within the \( \sigma \)-algebra remains within it as well.
  • Closed under complements: If a set is in the \( \sigma \)-algebra, its complement is also included. This means if one set belongs, so does everything outside it in the universe considered.
  • Contains the empty set: By definition, every \( \sigma \)-algebra will include the empty set, as it acts as a basic element of the space.
These properties ensure that a \( \sigma \)-algebra is stable under these operations, making it an ideal candidate for defining and working with measures. Exploring these properties lays the foundation for understanding more complex structures in mathematical analysis.

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Most popular questions from this chapter

Let \(\mathcal{M}\) be an infinite \(\sigma\)-algebra. a. M contains an infinite sequence of disjoint sets. b. \(\operatorname{card}(\mathcal{M}) \geq c\).

Let \((X, \mathcal{N}, \mu)\) be a measure space. A set \(E \subset X\) is called locally measurable if \(E \cap A \in \mathcal{M}\) for all \(A \in \mathcal{M}\) such that \(\mu(A)<\infty\). Let \(\widetilde{\mathcal{M}}\) be the collection of all locally measurable sets. Clearly \(\mathcal{M} \subset \widetilde{\mathcal{M}}\); if \(\mathcal{M}=\widetilde{\mathcal{M}}\), then \(\mu\) is called saturated. a. If \(\mu\) is \(\sigma\)-finite, then \(\mu\) is saturated. b. \(\widetilde{\mathcal{M}}\) is a \(\sigma\)-algebra. c. Define \(\tilde{\mu}\) on \(\tilde{\mathcal{M}}\) by \(\tilde{\mu}(E)=\mu(E)\) if \(E \in \mathcal{M}\) and \(\tilde{\mu}(E)=\infty\) otherwise. Then \(\widetilde{\mu}\) is a saturated measure on \(\tilde{\mathcal{M}}\), called the saturation of \(\mu\). d. If \(\mu\) is complete, so is \(\tilde{\mu}\). e. Suppose that \(\mu\) is semifinite. For \(E \in \widetilde{\mathcal{M}}\), define \(\mu(E)=\sup \\{\mu(A): A \in\) \(\mathcal{M}\) and \(A \subset E\\}\). Then \(\mu\) is a saturated measure on \(\tilde{\mathcal{M}}\) that extends \(\mu\). f. Let \(X_{1}, X_{2}\) be disjoint uncountable sets, \(X=X_{1} \cup X_{2}\), and \(\mathcal{M}\) the \(\sigma\)-algebra of countable or co-countable sets in \(X\). Let \(\mu_{0}\) be counting measure on \(\mathcal{P}\left(X_{1}\right)\), and define \(\mu\) on \(\mathcal{M}\) by \(\mu(E)=\mu_{0}\left(E \cap X_{1}\right)\). Then \(\mu\) is a measure on \(\mathcal{M}\), \(\widetilde{\mathcal{M}}=\mathcal{P}(X)\), and in the notation of parts (c) and (e), \(\widetilde{\mu} \neq \underline{\mu}\).

If \((X, \mathcal{M}, \mu)\) is a measure space and \(\left\\{E_{j}\right\\}_{1}^{\infty} \subset \mathcal{M}\), then \(\mu\left(\lim \inf E_{j}\right) \leq\) \(\liminf \mu\left(E_{j}\right) .\) Also, \(\mu\left(\lim \sup E_{j}\right) \geq \lim \sup \mu\left(E_{j}\right)\) provided that \(\mu\left(\bigcup_{1}^{\infty} E_{j}\right)<\) \(\infty\).

Let \((X, \mathcal{M}, \mu)\) be a finite measure space. a. If \(E, F \in \mathcal{M}\) and \(\mu(E \Delta F)=0\), then \(\mu(E)=\mu(F)\). b. Say that \(E \sim F\) if \(\mu(E \Delta F)=0\); then \(\sim\) is an equivalence relation on \(\mathcal{M}\). c. For \(E, F \in \mathcal{M}\), define \(\rho(E, F)=\mu(E \Delta F)\). Then \(\rho(E, G) \leq \rho(E, F)+\) \(\rho(F, G)\), and hence \(\rho\) defines a metric on the space \(\mathcal{M} / \sim\) of equivalence classes.

Let \(\mu\) be a finite measure on \((X, \mathcal{M})\), and let \(\mu^{*}\) be the outer measure induced by \(\mu\). Suppose that \(E \subset X\) satisfies \(\mu^{*}(E)=\mu^{*}(X)\) (but not that \(E \in \mathcal{M}\) ). a. If \(A, B \in \mathcal{M}\) and \(A \cap E=B \cap E\), then \(\mu(A)=\mu(B)\). b. Let \(\mathcal{M}_{E}=\\{A \cap E: A \in \mathcal{M}\\}\), and define the function \(\nu\) on \(\mathcal{M}_{E}\) defined by \(\nu(A \cap E)=\mu(A)\) (which makes sense by (a)). Then \(\mathcal{M}_{E}\) is a \(\sigma\)-algebra on \(E\) and \(\nu\) is a measure on \(\mathrm{M}_{E}\).

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