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

Short Answer

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\( \mathcal{M} \) contains an infinite sequence of disjoint sets, and \( \operatorname{card}(\mathcal{M}) \geq c \).

Step by step solution

01

Understand the Definition of a σ-algebra

A σ-algebra \( \mathcal{M} \) is a collection of sets that is closed under countable unions, countable intersections, and complements. Furthermore, it contains the empty set. The infinite nature of \( \mathcal{M} \) means it contains infinitely many sets.
02

Disjoint Sets in a σ-algebra

Since \( \mathcal{M} \) is a σ-algebra, it must include \( \emptyset \) and possibly an infinite set \( A \). Consider that \( A \) is partitioned into disjoint sets \( A_1, A_2, A_3, \ldots \). The \( \sigma\)-algebra being infinite implies that it can accommodate such partitions, as these partitions represent disjoint sets in \( \mathcal{M} \).
03

Cardinality of an Infinite σ-algebra

The cardinality of a σ-algebra, \( \operatorname{card}(\mathcal{M}) \), is at least continuum \( c \). This arises from the fact that a σ-algebra over an infinite set (such as the real numbers) includes all Borel sets, which form a set of cardinality \( c \). Since \( \mathcal{M} \) is infinite, the cardinality must satisfy \( \operatorname{card}(\mathcal{M}) \geq c \).
04

Conclude the Argument

By defining such disjoint sets \( A_1, A_2, \ldots \), the condition \( a \) is satisfied. Similarly, for condition \( b \), the cardinality of the σ-algebra \( \operatorname{card}(\mathcal{M}) \geq c \) follows directly from its nature of encompassing infinite constructs like power sets. Hence, both conditions in the problem are met.

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Key Concepts

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

Infinite Sequence of Disjoint Sets
An infinite sequence of disjoint sets is a collection of sets where none of the sets overlap with each other, and there are infinitely many of them. In simpler terms, no two sets share any common elements. This is an important concept when discussing a \(\sigma\)-algebra because one of the basic properties of such an algebra is its ability to manage countable unions and intersections of sets.

When we say a \(\sigma\)-algebra \( \mathcal{M} \) contains an infinite sequence of disjoint sets, it implies that you can take an infinite set \( A \) in the algebra and partition it into non-overlapping subsets \( A_1, A_2, A_3, \ldots \).

- **Partitioning**: To partition a set means to divide it into distinct parts that cover the entire original set.
- **Disjoint Sets**: Each set does not share elements with any other set in the sequence.

By having these properties, the \(\sigma\)-algebra can accommodate complex functions and probability spaces, providing a framework that is essential in many areas of mathematics such as measure theory.
Cardinality
Cardinality, in mathematical terms, is a measure of the "number of elements" in a set. For a \(\sigma\)-algebra \( \mathcal{M} \) that is infinite, we can talk about its cardinality to understand how vast it is in terms of element composition.

The size of such an algebra is described by the symbol \( c \), called the continuum, which represents the cardinality of the real numbers. In mathematical contexts:

- If a \(\sigma\)-algebra includes all Borel sets of the real line, it must have at least this cardinality \((c)\) because it encompasses every possible open set and their countable operations.
- Representing higher-order infinities, \(c\) exceeds traditional countable infinity \((\aleph_0)\) such as that of natural numbers.

Understanding cardinality in the context of a \(\sigma\)-algebra helps us grasp the "size" of these mathematical constructs, which in turn illuminates their capacity to handle and express complex sets and functions.
Borel Sets
Borel sets are a fundamental concept in measure theory. These sets are generated from the open intervals of real numbers through a process called \(\sigma\)-algebra generation. Starting from all open sets on the real line, through countable unions, intersections, and complements, we obtain the Borel sets.

Here's why Borel sets are significant:

- They provide a way to systematically include all kinds of subsets of real numbers in the framework of a \(\sigma\)-algebra.
- Borel sets include **all possible open sets**, and by extension through their properties, cover closed sets and other complex subsets.

In probability theory, Borel \(\sigma\)-algebras encompass events that can occur in a real-valued random process.
  • **Real Line Coverage**: Every subset of real numbers we might naturally encounter can be a Borel set.
  • **Versatility**: Borel sets make up a vital component of defining measurable spaces where complex functions are relevant.

The inclusion of Borel sets gives a \(\sigma\)-algebra a structured yet flexible way to manage continuity and limits, key elements in many mathematical analyses.

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Most popular questions from this chapter

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 \(\mathrm{M}_{E}\).

If \(\mu_{1}, \ldots, \mu_{n}\) are measures on \((X, \mathcal{M})\) and \(a_{1}, \ldots, a_{n} \in[0, \infty)\), then \(\sum_{1}^{n} a_{j} \mu_{j}\) is a measure on \((X, \mathcal{M})\).

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 \((X, \mathcal{N}, \mu)\) 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 \(\mu^{*}\) is an outer measure on \(X\) and \(\left\\{A_{j}\right\\}_{1}^{\infty}\) is a sequence of disjoint \(\mu^{*}\) measurable sets, then \(\mu^{*}\left(E \cap\left(\bigcup_{1}^{\infty} A_{j}\right)\right)=\sum_{1}^{\infty} \mu^{*}\left(E \cap A_{j}\right)\) for any \(E \subset X .\)

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