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Chapter 2: Problem 11

Suppose \(\mathcal{T}\) is a \(\sigma\) -algebra on a set \(Y\) and \(X \in \mathcal{T}\). Let \(\mathcal{S}=\\{E \in \mathcal{T}: E \subset X\\}\). (a) Show that \(\mathcal{S}=\\{F \cap X: F \in \mathcal{T}\\}\). (b) Show that \(\mathcal{S}\) is a \(\sigma\) -algebra on \(X\).

Short Answer

Expert verified
In this exercise, we proved that for a given σ-algebra on a set Y and a set X belonging to the σ-algebra, the set S containing elements from the σ-algebra that are subsets of X is equal to the set of intersections between the elements of the σ-algebra and X, i.e., S = {F ∩ X: F ∈ σ-algebra}. We also showed that S is a σ-algebra on X by verifying the three properties of σ-algebras.

Step by step solution

01

(Part a: Proving S = {F ∩ X: F ∈ σ-algebra})

First, we will show that S is a subset of {F ∩ X: F ∈ σ-algebra}. Let E be an arbitrary element of S, that is, E ∈ S. By definition of S, E ∈ σ-algebra and E ⊆ X. Since E ⊆ X, we can write E = E ∩ X, which shows that E is an intersection of an element of the σ-algebra and X. Therefore, E ∈ {F ∩ X: F ∈ σ-algebra}. Since E was arbitrary, we conclude that S ⊆ {F ∩ X: F ∈ σ-algebra}. Now, we will show that {F ∩ X: F ∈ σ-algebra} is a subset of S. Let G be an arbitrary element of the set {F ∩ X: F ∈ σ-algebra}. By definition, there exists an F ∈ σ-algebra such that G = F ∩ X. As F ∈ σ-algebra and X ∈ σ-algebra (given), their intersection G = F ∩ X is part of σ-algebra (property of σ-algebras). Furthermore, since G ⊆ X (as G is the intersection of F and X), G ∈ S. Since G was arbitrary, we conclude that {F ∩ X: F ∈ σ-algebra} ⊆ S. By proving both directions of the subset relationship, we can say that S = {F ∩ X: F ∈ σ-algebra}.
02

(Part b: Proving S is a σ-algebra on X)

Now, we need to show that S is a σ-algebra on X. To do this, we need to verify the following properties for S: 1. X ∈ S 2. If E ∈ S, then its complement with respect to X, denoted by E^c_X, is in S. 3. If \(E_1, E_2, \dots\) are elements of S, then the countable union of these sets, ∪_{i=1}^∞ E_i, is in S. 1. By the definition of S (elements of σ-algebra that are subsets of X), X is in S since X ⊆ X and given that X ∈ σ-algebra. 2. Let E ∈ S, which means E ∈ σ-algebra and E ⊆ X. We know that E ∩ X = E (since E ⊆ X), and using the result from part (a), we can write E = F ∩ X for some F ∈ σ-algebra. Then, the complement of E with respect to X is given by: \(E^c_X = X - E = X - (F ∩ X) = X ∩ (F^c)\) By the property of σ-algebras, the complement F^c also belongs to σ-algebra, and since \(E^c_X = X ∩ (F^c)\), \(E^c_X\) is an intersection of an element of σ-algebra and X, which implies that \(E^c_X ∈ S\). 3. Let \(E_1, E_2, \dots\) ∈ S. Then, these sets are also elements of σ-algebra and are subsets of X. By the property of σ-algebras: ∪_{i=1}^∞ E_i ∈ σ-algebra By the definition of S and since X ∈ σ-algebra, we know that X ⊆ S and thus, ∀i, \(E_i ⊆ X\). Therefore, the countable union of these sets also has elements only from X: ∪_{i=1}^∞ E_i ⊆ X Then, the countable union ∪_{i=1}^∞ E_i is an element of σ-algebra and a subset of X, which implies that ∪_{i=1}^∞ E_i ∈ S. Since S satisfies all three properties, we can conclude that S is a σ-algebra on X.

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Key Concepts

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

intersection of sets
When we talk about the intersection of sets, we refer to a fundamental operation in set theory. The intersection between two sets, say set **A** and set **B**, contains all the elements that are common to both sets. It’s like asking: "What do these sets have in common?" Represented symbolically, the intersection of **A** and **B** is noted as **A ∩ B**.

In the context of a sigma-algebra, which is a collection of sets closed under certain operations, the concept of intersections plays a crucial role. Sigma-algebras are particularly useful in probability and measure theory. For any two sets within the sigma-algebra, their intersection will also be part of the sigma-algebra. This property ensures that the sigma-algebra remains intact whenever you perform an intersection between its elements.
  • Theorem: If **F** and **X** are in a sigma-algebra, then **F ∩ X** is also in the sigma-algebra.
  • Application: This concept ensures that when two sets within a sigma-algebra are "intersected", the result is still within our familiar framework, adhering to pre-defined rules.
Hence, intersections aren't just about overlap; they are about maintaining structural integrity within mathematical frameworks like sigma-algebras.
subset
A subset is a fundamental concept in set theory where a set, say **A**, is considered a subset of another set, **B**, if all elements of **A** are also in **B**. We denote this relationship as **A ⊆ B**.

In the exercise context, the collection **\(\mathcal{S}\)** is defined as subsets of **X** found within a sigma-algebra **\(\mathcal{T}\)**. This means every set in **\(\mathcal{S}\)** is entirely contained within **X**. Saying that a set is a subset is like indicating its 'membership' to a broader set. For the sigma-algebra on **X**, the property that all elements are subsets of **X** is foundational, ensuring that any operations (like unions and complements) are contained within **X** itself.
  • Definition: For **E** to be an element of **\(\mathcal{S}\)**, we must have **E ⊆ X**.
  • Importance: Understanding subsets helps us grasp how collections like sigma-algebras are structured and maintained through set operations.
This keeps everything "within bounds" so to speak, ensuring all sigma-algebra properties are applied consistently.
set complement
The concept of a set complement is an essential operation in set theory. Given a set **A** within a universal set **U**, the complement, denoted as **Ac** or **U - A**, consists of all elements in **U** that are not in **A**. This operation effectively "inverts" the membership of the elements.

In the sigma-algebra context, if **E** is a set within a sigma-algebra over a space **Y**, its complement, determined relative to **X** (or the space over which the sigma-algebra is being examined), is significant. When dealing with complements, the crucial aspect is that the complement of a set in a sigma-algebra is also an element of this sigma-algebra.
  • Operation: Complement of **E** within **X** is denoted **Ec_X = X - E**.
  • Significance: Ensures the closure properties of sigma-algebras, maintaining all derived sets within the original space.
This property allows flexibility in manipulating sets while ensuring they remain within the structure of the sigma-algebra, making operations coherent.
countable union
A countable union refers to the union of a sequence of sets that can be indexed by the natural numbers. For instance, if you have sets **E1, E2, E3,...**, their countable union, symbolized as **\(\bigcup_{i=1}^{\infty} E_i\)**, includes all elements that appear in any of these sets.

In the context of sigma-algebras, a critical property is closure with respect to countable unions. This means that if all sets in a sequence belong to a sigma-algebra, their union also belongs to the same sigma-algebra. Such properties enable us to generate more complex sets from simpler ones while remaining within the established framework.
  • Property: If each **Ei** is in a sigma-algebra, then **\(\bigcup_{i=1}^{\infty} E_i\)** must also be within it.
  • Application: Countable unions allow expansion of datasets or event spaces without losing defined characteristics, maintaining mathematical consistency.
This ensures not only completeness but also mathematical robustness, adhering to specified rules of a sigma-algebra.

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Most popular questions from this chapter

Suppose \(r_{1}, r_{2}, \ldots\) is a sequence that contains every rational number. Let $$ F=\mathbf{R} \backslash \bigcup_{k=1}^{\infty}\left(r_{k}-\frac{1}{2^{k}}, r_{k}+\frac{1}{2^{k}}\right) $$ (a) Show that \(F\) is a closed subset of \(\mathbf{R}\). (b) Prove that if \(I\) is an interval contained in \(F,\) then \(I\) contains at most one element. (c) Prove that \(|F|=\infty\).

(a) Show that the set consisting of those numbers in (0,1) that have a decimal expansion containing one hundred consecutive 4 s is a Borel subset of \(\mathbf{R}\). (b) What is the Lebesgue measure of the set in part (a)?

Suppose \(B\) is a Borel set and \(f: B \rightarrow \mathbf{R}\) is a Lebesgue measurable function. Show that there exists a Borel measurable function \(g: B \rightarrow \mathbf{R}\) such that $$ |\\{x \in B: g(x) \neq f(x)\\}|=0 $$

Prove or give a counterexample: If \((X, \mathcal{S})\) is a measurable space and $$ f: X \rightarrow[-\infty, \infty] $$ is a function such that \(f^{-1}((a, \infty)) \in \mathcal{S}\) for every \(a \in \mathbf{R},\) then \(f\) is an \(\mathcal{S}\) -measurable function.

Prove that every open interval of \(\mathbf{R}\) contains either infinitely many or no elements in the Cantor set.
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