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Let \(X=Y\) be an uncountable linearly ordered set such that for each \(x \in X\), \(\\{y \in X: y

Short Answer

Expert verified
The integrals exist, but \(\iint \chi_{E} \, d \mu \, d \nu = 0\) and \(\iint \chi_{E} \, d \nu \, d \mu = 1\); they are not equal.

Step by step solution

01

Understand the Set and Measure

We have an uncountable linearly ordered set \(X=Y\) such that for each \(x \in X\), the set \(\{y \in X: y<x\}\) is countable. The \(\sigma\)-algebra \(\mathcal{M}=\mathcal{N}\) consists of sets that are countable or co-countable, and measure \(\mu=u\) assigns 0 to countable sets and 1 to co-countable sets.
02

Examine Set E and its Sections

Set \(E=\{(x, y) \in X \times X: y<x\}\) collects pairs \((x, y)\) such that \(y\) is less than \(x\). For any fixed \(x\), \(E_x = \{y \in X: y < x\}\) is countable. For any fixed \(y\), \(E^y = \{x \in X: y < x\}\) is co-countable.
03

Measurability of Sections

Since \(E_x\) is countable for all \(x\) and belongs to the \(\sigma\)-algebra of countable sets, \(E_x\) is measurable. Similarly, \(E^y\) is co-countable and belongs to the \(\sigma\)-algebra composing of co-countable sets, making \(E^y\) measurable.
04

Double Integrals and Fubini's Theorem

Consider the double integrals \(\iint \chi_{E} \, d \mu \, d u\) and \(\iint \chi_{E} \, d u \, d \mu\). By Fubini’s Theorem, these integrals exist because \(E_x\) and \(E^y\) are measurable for all \(x\) and \(y\).
05

Evaluating the Integrals

The integral \(\iint \chi_{E} \, d \mu \, d u\) sums over all \(x \in X\) the measure \(\mu(E_x)\), which equals 0 because each \(E_x\) is countable. Conversely, the integral \(\iint \chi_{E} \, d u \, d \mu\) sums over all \(y \in X\) the measure \(u(E^y)\), which equals 1 since each \(E^y\) is co-countable.

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Key Concepts

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

Uncountable Sets
In mathematics, an uncountable set is larger than any set that can be listed or counted one-by-one, such as the set of natural numbers. An example of an uncountable set is the set of real numbers. In the given exercise, the set \(X\) is uncountable, meaning there are more elements than can be mapped to the natural numbers. Yet interestingly, for any element \(x\) in \(X\), the subset \(\{y \in X: y < x\}\) is countable. This creates a situation where, individually, elements lead to countable subsets, yet the whole remains uncountable. This concept helps explain many complex structures in measure theory, which often deals with managing and understanding different sizes of infinity and their implications.
Sigma-Algebra
A sigma-algebra \(\mathcal{M}\) is a collection of sets closed under complementation and countable unions. This structure forms the foundational framework for modern measure theory, allowing us to define measures on these collections of sets. In this context, \(\mathcal{M}\) is made up of sets that are either countable or co-countable (complements of countable sets). The sigma-algebra enables us to talk about measurability and orthogonality, such as determining which sets can be measured with respect to a given measure \(\mu\), like in the problem where \(\mu\) assigns zero to countable sets and one to co-countable sets. Understanding sigma-algebra is crucial for handling abstractions in mathematical analysis and providing a way to explore the concept of 'size' or 'measure' in more depth.
Measurability
Measurability is an essential concept that connects sigma-algebras and measures, allowing us to discuss which sets can be assigned a measure, and how. For a set to be considered measurable, it must be able to exist within a given sigma-algebra. In our exercise, \(E_x\) and \(E^y\) are measurable because they are part of \(\mathcal{M}\), the sigma-algebra of countable and co-countable sets. The measurability of a set implies it can be integrated with respect to a given measure, which then leads into calculating things like our double integrals \(\iint \chi_{E} d \mu d u\), where \(\chi_{E}\) represents the characteristic function of \(E\). This forms the basis for assessing probabilities, volumes, and other areas needing precise measurement.
Fubini's Theorem
Fubini's Theorem is a pivotal result in integration theory that allows us to compute double integrals through iterated integrals, often simplifying complex multivariable integration. The theorem states that if the integrals of the absolute value of a function over the respective spaces are finite, then the iterated integrals are equal and the double integral can be reversed. In the exercise problem, examining \(\iint \chi_{E} \, d \mu \, d u\) and \(\iint \chi_{E} \, d u \, d \mu\) reveals that although the double integrals exist, they are not necessarily equal. This is due to the nature of \(E_x\) being countable and \(E^y\) being co-countable, reflecting a special circumstance where the order of integration impacts the result. Understanding Fubini's Theorem is crucial for tackling real-world problems that extend into higher dimensions and require comprehensive integration techniques.

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Most popular questions from this chapter

Suppose \(f, g: X \rightarrow \mathbb{R}\) are measurable. a. \(f g\) is measurable (where \(0 \cdot(\pm \infty)=0\) ). b. Fix \(a \in \overline{\mathbb{R}}\) and define \(h(x)=a\) if \(f(x)=-g(x)=\pm \infty\) and \(h(x)=\) \(f(x)+g(x)\) otherwise. Then \(h\) is measurable.

If \(\mu\left(E_{n}\right)<\infty\) for \(n \in \mathbb{N}\) and \(\chi_{E_{n}} \rightarrow f\) in \(L^{1}\), then \(f\) is (a.e. equal to) the characteristic function of a measurable set.

Let \(f(x)=x^{-1} \sin x\). a. Show that \(\int_{0}^{\infty}|f(x)| d x=\infty\). b. Show that \(\lim _{b \rightarrow \infty} \int_{0}^{b} f(x) d x=\frac{1}{2} \pi\) by integrating \(e^{-x y} \sin x\) with respect to \(x\) and \(y\). (In view of part (a), some care is needed in passing to the limit as \(b \rightarrow \infty .)\)

Suppose that \(f\) is a function on \(\mathbb{R} \times \mathbb{R}^{k}\) such that \(f(x, \cdot)\) is Borel measurable for each \(x \in \mathbb{R}\) and \(f(\cdot, y)\) is continuous for each \(y \in \mathbb{R}^{k}\). For \(n \in \mathbb{N}\), define \(f_{n}\) as follows. For \(i \in \mathbb{Z}\) let \(a_{i}=i / n\), and for \(a_{i} \leq x \leq a_{i+1}\) let $$ f_{n}(x, y)=\frac{f\left(a_{i+1}, y\right)\left(x-a_{i}\right)-f\left(a_{i}, y\right)\left(x-a_{i+1}\right)}{a_{i+1}-a_{i}} $$ Then \(f_{n}\) is Borel measurable on \(\mathbb{R} \times \mathbb{R}^{k}\) and \(f_{n} \rightarrow f\) pointwise; hence \(f\) is Borel measurable on \(\mathbb{R} \times \mathbb{R}^{k}\). Conclude by induction that every function on \(\mathbb{R}^{n}\) that is continuous in each variable separately is Borel measurable.

Suppose \((X, \mathcal{M}, \mu)\) is a \(\sigma\)-finite measure space and \(f \in L^{+}(X) .\) Let $$ G_{f}=\\{(x, y) \in X \times[0, \infty]: y \leq f(x)\\} $$ Then \(G_{f}\) is \(\mathrm{M} \times \mathrm{B}_{\mathrm{R}}\)-measurable and \(\mu \times m\left(G_{f}\right)=\int f d \mu ;\) the same is also true if the inequality \(y \leq f(x)\) in the definition of \(G_{f}\) is replaced by \(y

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