/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Let \(f=\left(f_{1}, f_{2}\right... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 1

Let \(f=\left(f_{1}, f_{2}\right): \Omega \rightarrow E \times F\). Show that \(f:(\Omega, \mathcal{A}) \rightarrow(E \times F ; \mathcal{E} \otimes \mathcal{F})\) is measurable if and only if \(f_{1}\) is measurable from \((\Omega, \mathcal{A})\) to \((E, \mathcal{E})\) and \(f_{2}\) is measurable from \((\Omega, \mathcal{A})\) to \((F, \mathcal{F})\)

Short Answer

Expert verified
The function \( f \) is measurable if and only if both \( f_1 \) and \( f_2 \) are measurable.

Step by step solution

01

Understanding the Problem

We are given a mapping \( f = (f_1, f_2) \) where \( f: \Omega \rightarrow E \times F \). The objective is to show that \( f \) is measurable with respect to the product \( \sigma\)-algebra \( \mathcal{E} \otimes \mathcal{F} \) if and only if \( f_1 \) and \( f_2 \) are measurable with respect to their respective \( \sigma\)-algebras.
02

Definition of Measurability

A function is measurable if the preimage of any measurable set in the codomain is measurable in the domain. Here, \( f \) is measurable if for every measurable set \( A \in \mathcal{E} \otimes \mathcal{F} \), the preimage \( f^{-1}(A) \) belongs to \( \mathcal{A} \). Similarly, \( f_1 \) and \( f_2 \) are measurable if \( f_1^{-1}(A') \in \mathcal{A} \) for every \( A' \in \mathcal{E} \) and \( f_2^{-1}(B') \in \mathcal{A} \) for every \( B' \in \mathcal{F} \).
03

Measurability of the Product Function

The product \( \sigma\)-algebra \( \mathcal{E} \otimes \mathcal{F} \) is generated by measurable rectangles, i.e., sets of the form \( A' \times B' \) where \( A' \in \mathcal{E} \) and \( B' \in \mathcal{F} \). Thus, \( f \) is measurable if \( f^{-1}(A' \times B') \in \mathcal{A} \) for all measurable rectangles.
04

Showing \( f \) Measurable Implies \( f_1, f_2 \) Measurable

Consider rectangles \( A' \times F \) and \( E \times B' \). The preimage \( f^{-1}(A' \times F) = f_1^{-1}(A') \), since \( f_1(x) \) must be in \( A' \) and there are no conditions on \( f_2 \). Hence \( f^{-1}(A' \times F) \in \mathcal{A} \), establishing \( f_1 \) is measurable. Similarly, \( f^{-1}(E \times B') = f_2^{-1}(B') \), proving \( f_2 \) is measurable.
05

Showing \( f_1, f_2 \) Measurable Implies \( f \) Measurable

Assuming \( f_1 \) and \( f_2 \) are measurable, we need to show \( f^{-1}(A' \times B') = f_1^{-1}(A') \cap f_2^{-1}(B') \in \mathcal{A} \). Since intersections of measurable spaces are measurable, \( f^{-1}(A' \times B') \) is indeed measurable. Hence \( f \) is measurable.
06

Conclusion

We have shown both directions: a measurable product function implies its components are measurable, and measurable components imply the product function is measurable. Thus, the statement holds.

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Key Concepts

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

Product Sigma-Algebra
To understand how product sigma-algebras work, let's first grasp the basic idea of a sigma-algebra. A sigma-algebra is a collection of sets that is closed under countable unions, intersections, and complements. This structure is foundational in measure theory because it allows us to define measures on these sets.
  • The product sigma-algebra, denoted as \( \mathcal{E} \otimes \mathcal{F} \), is constructed from two individual sigma-algebras, \( \mathcal{E} \) on set \( E \) and \( \mathcal{F} \) on set \( F \).
  • It is generated by "measurable rectangles," which are simply products of measurable sets from \( \mathcal{E} \) and \( \mathcal{F} \). That means if \( A' \in \mathcal{E} \) and \( B' \in \mathcal{F} \), then \( A' \times B' \) belongs to \( \mathcal{E} \otimes \mathcal{F} \).
This structure allows us to explore more complex measures on multi-dimensional spaces by leveraging known measures on lower dimensions. Let's see how it ties into measurability in functions.
Preimage Measurability
Measurability of a function is a crucial concept in measure theory, requiring that every measurable set in the codomain leads back to a measurable set in the domain when you apply the function's inverse. This is known as preimage measurability.
  • If you have a function \( f : \Omega \to E \times F \), measurability means that for any measurable set \( A \) in the product sigma-algebra \( \mathcal{E} \otimes \mathcal{F} \), the preimage \( f^{-1}(A) \) belongs to the sigma-algebra \( \mathcal{A} \).
  • For individual components, \( f_1\) is measurable from \( \Omega \) to \( E \) if for any measurable set \( A' \in \mathcal{E} \), \( f_1^{-1}(A') \) is in \( \mathcal{A} \). Similarly, \( f_2\) is measurable if \( f_2^{-1}(B') \) is in \( \mathcal{A} \) for any \( B' \in \mathcal{F} \).
This ensures that operations done with the function in the context of measure theory maintain their measurable nature, which is especially important when using these functions in problems involving statistics or probability.
Measurable Rectangles
Measurable rectangles form the cornerstone of the product sigma-algebra. A measurable rectangle is simply the Cartesian product of two measurable sets, one from each of the constituent sigma-algebras.
  • Consider sets \( A' \in \mathcal{E} \) and \( B' \in \mathcal{F} \). Their product, \( A' \times B' \), is called a measurable rectangle because it represents a set of all element pairs \((e, f)\) where \( e \in A' \) and \( f \in B' \).
  • These rectangles generate the product sigma-algebra \( \mathcal{E} \otimes \mathcal{F} \), meaning that any set within this sigma-algebra can be "built" from these rectangles through basic operations like countable unions and intersections.
Grasping this concept helps in understanding how measurability of a function like \( f = (f_1, f_2) \), defined on a product space, can be tackled by analyzing each component function's interaction with these measurable rectangles. This is essential for proving deeper results in measure theory and integration over product spaces.

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Most popular questions from this chapter

Let \(A_{n}\) be a sequence of events. Show that $$ P\left(A_{n} \text { i.o. }\right) \geq \lim _{n \rightarrow \infty} \sup _{\infty} P\left(A_{n}\right) $$

Let \(X, Y\) be independent random variables taking values in \(\mathbf{N}\) with $$ P(X=i)=P(Y=i)=\frac{1}{2^{i}} \quad(i=1,2, \ldots) $$ Find the following probabilities: a) \(P(\min (X, Y) \leq i) \quad\left[\right.\) Ans.: \(\left.1-\frac{1}{4^{4}}\right]\) b) \(P(X=Y)\) [Ans.: \(\frac{1}{3}\) ] c) \(P(Y>X) \quad\left[\right.\) Ans.: \(\left.\sum_{i \geq 0}^{3} \frac{1}{2^{2}\left(2^{i}-1\right)}\right]\) d) \(P(X\) divides \(Y)\) [Ans,: \(\frac{1}{3}\) ] e) \(P(X \geq k Y)\) for a given positive integer \(k\) [Ans,: \(\frac{1}{2^{1+k}-1}\) ]

Construct an example to show that \(E\\{X Y\\}=E\\{X\\} E\\{Y\\}\) does not imply in general that \(X\) and \(Y\) are independent r.v.'s (we assume \(X, Y\) and \(X Y\) are all in \(L^{1}\) ).

Let \(\Omega=\mathbf{R}\) and \(\mathcal{A}=\mathcal{B} .\) Let \(P\) be given by \(P(A)=\frac{1}{\sqrt{2 \pi}} \int 1_{A}(x) e^{-x^{2} / 2} d x\) Let \(X(x)=x\). Show that \(X\) has a normal distribution with parameters \(\mu=0\) and \(\sigma^{2}=1\).

Toss a coin with \(P\) (Heads) \(=p\) repeatedly. Let \(A_{k}\) be the event that \(k\) or more consecutive heads occurs amongst the tosses numbered \(2^{k}, 2^{k}+\) \(1, \ldots, 2^{k+1}-1 .\) Show that \(P\left(A_{k}\right.\) i.o. \()=1\) if \(p \geq \frac{1}{2}\) and \(P\left(A_{k}\right.\) i.o. \()=0\) if \(p<\frac{1}{2} .\)
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