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A family of sets \(\mathcal{R} \subset \mathcal{P}(X)\) is called a ring if it is closed under finite unions and differences (i.e., if \(E_{1}, \ldots, E_{n} \in \mathcal{R}\), then \(\bigcup_{1}^{n} E_{j} \in \mathcal{R}\), and if \(E, F \in \mathcal{R}\), then \(E \backslash F \in \mathcal{R}\) ). A ring that is closed under countable unions is called a \(\sigma\)-ring. a. Rings (resp. \(\sigma\)-rings) are closed under finite (resp. countable) intersections. b. If \(\mathcal{R}\) is a ring (resp. \(\sigma\)-ring), then \(\mathcal{R}\) is an algebra (resp. \(\sigma\)-algebra) iff \(X \in \mathcal{R}\). c. If \(\mathcal{R}\) is a \(\sigma\)-ring, then \(\left\\{E \subset X: E \in \mathcal{R}\right.\) or \(\left.E^{c} \in \mathcal{R}\right\\}\) is a \(\sigma\)-algebra. d. If \(\mathcal{R}\) is a \(\sigma\)-ring, then \(\\{E \subset X: E \cap F \in \mathcal{R}\) for all \(F \in \mathcal{R}\\}\) is a \(\sigma\)-algebra.

Short Answer

Expert verified
Rings and \( \sigma \)-rings are closed under intersections; \( X \in \mathcal{R} \) makes them algebras; specific collections form \( \sigma \)-algebras.

Step by step solution

01

Understanding Ring and σ-Ring

A ring \( \mathcal{R} \) is a collection of sets that is closed under finite unions and set differences. A \( \sigma \)-ring is a ring that is also closed under countable unions.
02

Proving Rings Are Closed Under Finite Intersections

To prove part (a), observe that for any sets \( E, F \in \mathcal{R} \), the intersection \( E \cap F \) can be expressed as a set difference from their union: \( E \cap F = E \setminus (E \setminus F) \). Since \( \mathcal{R} \) is closed under set differences and finite unions, it follows that \( E \cap F \in \mathcal{R} \). By induction, this property holds for any finite number of sets.
03

Proving \( \sigma \)-Rings Are Closed Under Countable Intersections

For \( \sigma \)-rings, consider sets \( E_1, E_2, \ldots \) such that each \( E_i \in \mathcal{R} \). The intersection \( \bigcap_{i=1}^{\infty} E_i \) is equal to \( X \setminus \bigcup_{i=1}^{\infty} (X \setminus E_i) \). Since \( \sigma \)-rings are closed under countable unions and complements, \( \bigcap_{i=1}^{\infty} E_i \in \mathcal{R} \). Hence, they are closed under countable intersections.
04

Connection Between Ring/σ-Ring and Algebra/σ-Algebra

For part (b), if \( X \in \mathcal{R} \), then for any \( E \in \mathcal{R} \), \( X \setminus E \in \mathcal{R} \) as it is a set difference. This implies closure under complements, making \( \mathcal{R} \) an algebra. Similarly, if \( \mathcal{R} \) is a \( \sigma \)-ring and \( X \in \mathcal{R} \), it is closed under all countable operations, thus a \( \sigma \)-algebra.
05

Forming a σ-Algebra from a σ-Ring

Part (c) involves checking that the collection \( \{ E \subset X : E \in \mathcal{R} \text{ or } E^c \in \mathcal{R} \} \) is a \( \sigma \)-algebra. This collection includes the whole set \( X \), is closed under complements by definition, and if \( \{ E_i \} \subset \mathcal{R} \), both \( \bigcup E_i \) and its complement are in the collection, ensuring closure under countable unions.
06

Proving the Condition Forms a σ-Algebra

For part (d), the collection \( \{ E \subset X : E \cap F \in \mathcal{R} \text{ for all } F \in \mathcal{R} \} \) now must satisfy the \( \sigma \)-algebra properties. It includes \( X \), is closed under countable intersections because each intersection with any \( F \) results in a set within \( \mathcal{R} \), and Handle complements by checking all \( E_i \cap F \).'

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Key Concepts

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

Ring of Sets
Imagine a ring of sets as a collection, or a family of sets, that allows us to perform specific operations while keeping the results within the same family. For a collection of sets \mathcal{R} to be called a ring, it must be closed under two significant operations: finite unions and set differences.
  • **Finite Unions:** This means that if you take any finite number of sets from the collection and unite them, the resulting set should also be in the collection. For example, if you have sets \(A\) and \(B\) in \mathcal{R}, their union \(A \cup B\) must be in \mathcal{R} too.
  • **Set Differences:** Similarly, if you take the difference between any two sets in the collection, that difference should also belong to the collection. So, if \(A\) and \(B\) are in \mathcal{R}, then \(A \setminus B\) should be in \mathcal{R}.
These operations ensure that the structure of the collection remains intact, no matter how you combine or subtract sets within it.
Sigma-Ring
A \(\sigma\)-ring builds on the concept of a ring by introducing more flexibility with the types of unions it can handle. Besides being closed under finite unions and set differences like a ring, a \(\sigma\)-ring can handle countable unions.
In simple terms, if \mathcal{R} is a \(\sigma\)-ring and you have an infinite sequence of sets \(E_1, E_2, E_3, \ldots\) within \mathcal{R}, their union \(\bigcup_{i=1}^{\infty} E_i\) must also be in \mathcal{R}.
  • This ability to manage countable unions makes \(\sigma\)-rings a more robust structure compared to rings, especially when dealing with infinite processes.
  • The closure under countable unions is crucial in measure theory, as it helps in defining measures on infinite collections of sets.
Thus, \(\sigma\)-rings are indispensable in advanced mathematical analyses involving infinite series and sequences.
Sigma-Algebra
Sigma-algebras are a step further in the hierarchy of set structures. A \(\sigma\)-algebra is a \(\sigma\)-ring that includes the entire universal set \(X\) in its collection. This one property makes a huge difference in function!
The additional properties imposed on \(\sigma\)-algebras are:
  • **Closure under Complements:** If any set \(E\) is in the \(\sigma\)-algebra, its complement \(E^c\) must also be present. \(\sigma\)-rings might not necessarily have this property unless the whole set is included, which turns them into \(\sigma\)-algebras.
  • **The Entire Set is Included:** Unlike \(\sigma\)-rings, which can lack the universal set, a \(\sigma\)-algebra always includes \(X\).
These properties are fundamental for measuring concepts in mathematics, as they allow us to treat the entire space and every possible complement, facilitating easier computations and study of measures.
Set Operations
Set operations form the backbone of understanding rings, \(\sigma\)-rings, and \(\sigma\)-algebras. Familiar operations include union, intersection, and set difference.
  • **Union (\cupg):** Combining all elements from two sets. If you have sets \(E\) and \(F\), their union includes every element from both.
  • **Intersection (\capg):** It contains only elements that are common to both sets. Interestingly, in rings and \(\sigma\)-rings, intersections can be achieved through the combination of unions and differences.
  • **Complement:** This operation relates to the whole set \(X\). The complement of a set \(E\) is all the elements not in \(E\).
  • **Difference (\setminusg):** The set difference \(E \setminus F\) includes elements present in \(E\) but not in \(F\).
Understanding these operations is crucial, as they define how we combine or separate sets within rings, \(\sigma\)-rings, and \(\sigma\)-algebras, impacting how we apply theories in broader mathematical problems.

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Most popular questions from this chapter

If \(E \in \mathcal{L}\) and \(m(E)>0\), for any \(\alpha<1\) there is an open interval \(I\) such that \(m(E \cap I)>\alpha m(I)\).

Let \(\mathcal{A} \subset \mathcal{P}(X)\) be an algebra, \(\mathcal{A}_{\sigma}\) the collection of countable unions of sets in \(\mathcal{A}\), and \(\mathcal{A}_{\sigma \sigma}\) the collection of countable intersections of sets in \(\mathcal{A}_{\sigma}\). Let \(\mu_{0}\) be a premeasure on \(\mathcal{A}\) and \(\mu^{*}\) the induced outer measure. a. For any \(E \subset X\) and \(\epsilon>0\) there exists \(A \in \mathcal{A}_{\sigma}\) with \(E \subset A\) and \(\mu^{*}(A) \leq\) \(\mu^{*}(E)+\epsilon\). b. If \(\mu^{*}(E)<\infty\), then \(E\) is \(\mu^{*}\)-measurable iff there exists \(B \in A_{\sigma \delta}\) with \(E \subset B\) and \(\mu^{*}(B \backslash E)=0\). c. If \(\mu_{0}\) is \(\sigma\)-finite, the restriction \(\mu^{*}(E)<\infty\) in (b) is superfluous.

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 \((X, \mathcal{M}, \mu)\) 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.

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 \(\mu\left(\bigcup_{1}^{\infty} E_{j}\right)<\) \(\infty\).

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