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Chapter 2: Problem 1

Let \(\mathcal{A}=\left\\{A \subset I \mid A\right.\) is countable, or \(A^{c}\) is countable \(\\}\). Prove that \(\mathcal{A}\) is a \(\sigma\)-algebra.

Short Answer

Expert verified
\(\mathcal{A}\) is a sigma-algebra as it satisfies all required properties.

Step by step solution

01

Understand the Definition of a Sigma-algebra

A sigma-algebra \(\mathcal{A}\) over a set \(I\) is a collection of subsets of \(I\) that satisfies three properties: (1) The empty set is in \(\mathcal{A}\), (2) If a set \(A\) is in \(\mathcal{A}\), then its complement \(A^c\) is also in \(\mathcal{A}\), and (3) \(\mathcal{A}\) is closed under countable unions.
02

Prove the Empty Set is in \(\mathcal{A}\)

The empty set \(\emptyset\) is countable, hence by the definition of \(\mathcal{A}\), \(\emptyset \in \mathcal{A}\). This satisfies the first property of a sigma-algebra.
03

Prove Complements are in \(\mathcal{A}\)

If \(A \in \mathcal{A}\), then either \(A\) is countable or \(A^c\) is countable. If \(A\) is countable, then \(A^c\) must be such that \((A^c)^c = A\) is countable; hence, by definition \(A^c\) will be in \(\mathcal{A}\). This satisfies the second property.
04

Prove Closure Under Countable Unions

Let \(A_1, A_2, A_3, \ldots\) be a sequence of sets in \(\mathcal{A}\). Each set is either countable or its complement is countable. The union \(\bigcup_{k=1}^{\infty} A_k\) must be countable if each \(A_k\) is countable. If not, consider assuming \(B = \bigcup_{k=1}^{\infty} A_k\) isn't countable. If this is the case, then \(B^c\), having potentially a countable complement from each \(A_k^c\) being countable due to a finite intersection or De Morgan's laws, especially when finite except one being infinite, will be countable.
05

Confirm \(\mathcal{A}\) Satisfies All Sigma-algebra Properties

Since \(\mathcal{A}\) contains the empty set, is closed under taking complements, and closed under countable unions, we can conclude \(\mathcal{A}\) is a sigma-algebra based on the fulfilled criteria.

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Key Concepts

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

Countable Sets
A countable set is one that can be put in a one-to-one correspondence with the natural numbers. This means we can list the elements of the set in a sequence like 1, 2, 3, and so on. Countable sets may be finite or infinite.

Here are some key features of countable sets:
  • Finite Sets: Any set with a limited number of elements is countable because you can count them one by one.
  • Infinite Sets: Some infinite sets like the set of all integers or rational numbers are countable. Although they are infinite, you can still list them without missing any.
Understanding countability is crucial when dealing with sigma-algebras, as they often involve operations including countable unions or intersections.
Complement of a Set
The complement of a set, denoted as \(A^c\), consists of all the elements not in set \(A\) relative to a universal set \(I\). If you have a set within a larger context, the complement is everything outside that set within the same context.

Here's why complements are important:
  • Understanding Relationships: Complements help define relationships between sets and their surroundings.
  • In Sigma-Algebras: If a set is in a sigma-algebra, its complement must also be included. This property helps maintain balance within the collection of sets.
Handling complements is foundational in proving whether a collection of sets is a sigma-algebra because they ensure closure under set complementation.
Countable Unions
Countable unions refer to the union of a sequence of sets that can be indexed by natural numbers like \(A_1, A_2, A_3, \ldots\). If these sets are part of a sigma-algebra, their union must also be a part of the same sigma-algebra.

In sigma-algebras, countable unions are important for a few reasons:
  • Closure Property: A sigma-algebra is closed under countable unions. This means the union of any number of sets from the sigma-algebra is also in the sigma-algebra.
  • Handling Infinite Collections: Even if you have an infinite sequence of sets, as long as they are countable, their union will still be part of the sigma-algebra.
Countable unions are a key characteristic that differentiates sigma-algebras from other types of collections, ensuring they can handle large or infinite combinations effectively.

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Most popular questions from this chapter

Prove that if \(X \subset I\) is measurable, then for any \(\varepsilon>0\) there is an open set \(U\) containing \(X\) such that \(\mu(U \backslash X)<\varepsilon\). This is sometimes referred to as the first of Littlewood's three principles.

Prove that if \(A \subset I=[0,1]\) has measure \(\mu(A)<1\) and \(\varepsilon>0\), then there is an interval \([a, b] \subset I\) such that \(\mu(A \cap[a, b])<\) \(\varepsilon(b-a)\).

Suppose \(\mathcal{A}_{\lambda}\) is a \(\sigma\)-algebra of subsets of \(X\) for each \(\lambda\) in some indexing set \(\Lambda\). Prove that $$ \mathcal{A}=\bigcap_{\lambda \in \Lambda} \mathcal{A}_{\lambda} $$ is a \(\sigma\)-algebra of subsets of \(X\).

Let \(A\) be a measurable set with \(\mu(A)>0\) and let $$ \Delta=\left\\{x_{1}-x_{2} \mid x_{1}, x_{2} \in A\right\\} $$ be the set of differences of elements of \(A\). Then for some \(\varepsilon>0\) the set \(\Delta\) contains the interval \((-\varepsilon, \varepsilon)\).

Prove that \(C\) is the subset of elements of \([0,1]\) which can be represented in base three using only the digits 0 and 2 . More precisely, prove that \(x \in C\) if and only if it can be expressed in the form $$ x=\sum_{n=1}^{\infty} \frac{c_{n}}{3^{n}} $$ where each \(c_{n}\) is either 0 or 2 .
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