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Chapter 2: Problem 12

Suppose \(b

Short Answer

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So, to prove that a subset \(A\) of the interval \((b, c)\) is Lebesgue measurable if and only if \(|A| + |(b, c) \backslash A| = c - b\), first, note that if \(A\) is Lebesgue measurable, then by definition, \(|A|\) is the infimum of the sums of the lengths of countable collections of disjoint open intervals that cover \(A\). Similarly, if \((b, c) \backslash A\) is Lebesgue measurable, then there exist disjoint open intervals that cover the complement set. Next, prove that if \(A\) is Lebesgue measurable, then \(|A| + |(b, c) \backslash A| = c - b\). This follows since their union covers the entire interval \((b, c)\), and the sum of their lengths is equal to \(c - b\). Conversely, if \(|A| + |(b, c) \backslash A| = c - b\), then both \(A\) and \((b, c) \backslash A\) satisfy the condition for Lebesgue measurability with positive 蔚, and hence \(A\) is Lebesgue measurable.

Step by step solution

01

Before we start, we need to understand what it means for a set to be Lebesgue measurable and how to calculate the Lebesgue measure of a set. A set A is Lebesgue measurable if for every 蔚 > 0, there exists a countable collection of disjoint open intervals {I_n} such that 1. A is a subset of the union of the intervals I_n; 2. the difference between the sum of the lengths of I_n and the outer measure of A is less than 蔚. The Lebesgue measure |A| of a measurable set A is defined as the infimum (greatest lower bound) of the sum of the lengths of any countable collection of open intervals that covers A. In our problem, we are given that A is a subset of (b, c), so we know that the outer measure of A is less than or equal to the Lebesgue measure of (b, c), which is equal to c - b. #Step 2: The complement (b, c) \ A#

Let's consider the complement of A within the given interval, denoted by (b, c) \ A. By definition, this complement contains all the points in (b, c) that are not in A. Notice that A and (b, c) \ A are disjoint and their union covers the entire interval (b, c). #Step 3: A is measurable if and only if (b, c) \ A is measurable#
02

We know that A is measurable if it satisfies the condition mentioned in Step 1. Similarly, (b, c) \ A is measurable if the same condition is satisfied for this complement set. Let 蔚 > 0. If A is Lebesgue measurable, then there exist disjoint open intervals {I_n} that cover A and satisfy the condition mentioned in Step 1. Similarly, if (b, c) \ A is Lebesgue measurable, then there exist disjoint open intervals {J_n} that cover (b, c) \ A and satisfy the same condition. #Step 4: Proof of the main result#

We want to prove that A is Lebesgue measurable if and only if the sum of the Lebesgue measures of A and (b, c) \ A is equal to the length c - b of the interval (b, c). (鈬�) Suppose A is Lebesgue measurable. Then, by definition, |A| is the infimum of the sum of the lengths of the countable collection of disjoint open intervals {I_n} that covers A. Since the open intervals {I_n} and {J_n} cover A and (b, c) \ A respectively, their union {I_n 鈭� J_n} covers the entire interval (b, c), and the sum of their lengths is equal to c - b. Therefore, |A| + |(b, c) \ A| = c - b. (鈬�) Suppose |A| + |(b, c) \ A| = c - b. We know that the outer measure of A is less than or equal to the Lebesgue measure of (b, c) and the outer measure of (b, c) \ A is less than or equal to the Lebesgue measure of (b, c). Therefore, the sum of the outer measures of A and (b, c) \ A is less than or equal to 2(c - b). Since |A| + |(b, c) \ A| = c - b, we have that A and (b, c) \ A are both Lebesgue measurable, because they satisfy the condition mentioned in Step 1 for 蔚 = (c - b)/2. Therefore, A is Lebesgue measurable. This completes the proof that A is Lebesgue measurable if and only if |A| + |(b, c) \ A| = c - b.

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Key Concepts

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

Lebesgue Measurable Sets
A Lebesgue measurable set is an essential concept in measure theory, which explores ways to assign sizes to sets where traditional measures might not work. Imagine Lebesgue measure as a more flexible ruler that can measure even bizarrely shaped sets where regular rulers fail.

For a set to be declared Lebesgue measurable, it must meet specific criteria involving open intervals, which I will explain. For any small number \( \varepsilon > 0 \), you must be able to find a collection of open intervals that satisfy two conditions:
  • The set must be fully contained within these intervals.
  • The total length of these intervals must not exceed the outer measure of the set by more than \( \varepsilon \).
This ensures that the approximation of the set's measure is as accurate as desired, a crucial quality for mathematical analysis.
Outer Measure
Outer measure is like a preliminary step in determining the true 'size' or measure of a set. It's a concept that helps us understand the set without initially worrying about complex behavior it might have.

In practical terms, the outer measure of a set is calculated as the smallest total length of open intervals that can cover the entire set. Open intervals are the usual building blocks in this context, representing straightforward segments of the real line. Although not always the perfect fit, they offer a way to estimate the extent of the set by encapsulating its entirety.

The outer measure technique is delicate, too. It gives us a bound, known as the infimum, without assuming the set is neat. It allows us to manage sets that lie outside conventional methods of measurement, making it an indispensable tool in measure theory.
Measure Theory
Measure theory is the area of mathematics that enables us to extend the notion of length, area, and volume beyond simple shapes. It significantly broadens the scope of what can be measured 鈥� imagine measuring a cloud or a path in the dark.

By focusing on measurable sets, measure theory helps mathematicians and scientists to quantify characteristics of spaces that aren't straightforwardly handled by classical geometry. It's the backbone of modern probability and integral calculus, providing a rigorous foundation allowing challenges to traditional intuitions.

At its core, measure theory requires defining and comparing collections of sets to understand how to measure 'complicated' objects. Through this, we identify properties like sigma-algebras, which define collections of sets that are closed under the complement and countable unions, connecting neatly to Lebesgue measurability.
Set Complement
Understanding the complement of a set is crucial for grasping relationships between sets. In the context of an interval, the complement captures everything that lies within the given bounds but not in the original set.

Consider an interval \((b, c)\) and a set \(A \subset (b, c)\). The complement, denoted by \((b, c) \backslash A\), includes all the points in \((b, c)\) that aren't in \(A\). Essentially, two contrasting puzzles pieces that perfectly fit to complete the interval. For measure theory, examining both a set and its complement helps understand the full landscape of the interval.

Since \(A\) and \((b, c) \backslash A\) are disjoint and their union equals the entire interval, the sum of their measures gives \(c-b\). This observation forms the core part of proving Lebesgue measurability and gives insights into how set complements interact with outer measures.

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Most popular questions from this chapter

Suppose \(\mathcal{S}\) is the smallest \(\sigma\) -algebra on \(\mathbf{R}\) containing \(\\{[r, \infty): r \in \mathbf{Q}\\}\). Prove that \(\mathcal{S}\) is the collection of Borel subsets of \(\mathbf{R}\).

Prove that the collection of Lebesgue measurable subsets of \(\mathbf{R}\) is translation invariant. More precisely, prove that if \(A \subset \mathbf{R}\) is Lebesgue measurable and \(t \in \mathbf{R},\) then \(t+A\) is Lebesgue measurable.

Suppose \(A \subset \mathbf{R}\). Prove that \(A\) is Lebesgue measurable if and only if $$ |(-n, n) \cap A|+|(-n, n) \backslash A|=2 n $$ for every \(n \in \mathbf{Z}^{+}\).

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 \(X\).

Suppose \(X\) is a finite set. Explain why a sequence of functions from \(X\) to \(\mathbf{R}\) that converges pointwise on \(X\) also converges uniformly on \(X\).
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