Chapter 1: Problem 3
Let \(X\) have a noncountable number of points. Set \(\mu^{*}(E)=0\) if \(E\) is countable, \(\mu^{*}(E)=1\) if \(E\) is noncountable. Show that \(\mu^{*}\) is an outer measure, and determine the measurable sets.
Short Answer
Expert verified
\( \mu^{*} \) is an outer measure, and all subsets of \( X \) are \( \mu^{*} \)-measurable.
Step by step solution
01
Understanding the Outer Measure Definition
An outer measure \( \mu^{*} \) on a set \( X \) is a function that assigns a non-negative number to every subset of \( X \) such that: 1. \( \mu^{*}(\emptyset) = 0 \)2. \( \mu^{*}(A) \leq \mu^{*}(B) \) if \( A \subseteq B \)3. \( \mu^{*}\left(\bigcup_{i=1}^{\infty} E_i\right) \leq \sum_{i=1}^{\infty} \mu^{*}(E_i) \) for any sequence \( \{E_i\} \) of subsets of \( X \).
02
Verifying Condition 1
Check that \( \mu^{*}(\emptyset) = 0 \): The empty set is countable, so by the definition of \( \mu^{*} \), \( \mu^{*}(\emptyset) = 0 \).
03
Verifying Condition 2
Check if \( \mu^{*}(A) \leq \mu^{*}(B) \) when \( A \subseteq B \):- If \( A \) is countable, then \( \mu^{*}(A) = 0 \).- If \( B \) is countable, then \( \mu^{*}(B) = 0 \) too, so \( \mu^{*}(A) \leq \mu^{*}(B) \) holds. - If \( B \) is noncountable, then \( \mu^{*}(B) = 1 \), but since \( A \subseteq B \), \( A \) cannot be noncountable when \( B \) is countable. Thus, if \( A \) is noncountable, \( \mu^{*}(A) = 1 \leq 1 = \mu^{*}(B) \).
04
Verifying Condition 3
Check if \( \mu^{*}\left(\bigcup_{i=1}^{\infty} E_i\right) \leq \sum_{i=1}^{\infty} \mu^{*}(E_i) \): - If \( \bigcup_{i=1}^{\infty} E_i \) is countable, then \( \mu^{*}\left(\bigcup_{i=1}^{\infty} E_i\right) = 0 \), which is less than or equal to any non-negative sum. - If \( \bigcup_{i=1}^{\infty} E_i \) is noncountable, at least one \( E_i \) must be noncountable (and hence \( \mu^{*}(E_i) = 1 \)). Thus, \( \mu^{*}\left(\bigcup_{i=1}^{\infty} E_i\right) = 1 \leq \sum_{i=1}^{\infty} \mu^{*}(E_i) \).
05
Determining the Measurable Sets
A set \( E \) is \( \mu^{*} \)-measurable if for every set \( A \subseteq X \), we have:\[ \mu^{*}(A) = \mu^{*}(A \cap E) + \mu^{*}(A \setminus E) \]- If \( E \) is countable, then for any set \( A \), since \( A \cap E \) is countable, \( \mu^{*}(A \cap E) = 0 \) and \( \mu^{*}(A \setminus E) = \mu^{*}(A) \).- If \( E \) is noncountable, \( A \cap E \) may be noncountable making \( \mu^{*}(A \cap E) = 1 \), and thus \( \mu^{*}(A \setminus E) = 0 \) which implies their sum equals \( \mu^{*}(A) \).Hence, all subsets of \( X \) are \( \mu^{*} \)-measurable.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Noncountable Sets
Understanding noncountable sets is crucial in measure theory. A noncountable set is an infinite set that cannot be counted, meaning its elements cannot be put into a one-to-one correspondence with the natural numbers. This is unlike countable sets, such as the set of integers, where each element can be counted in succession. An example of a noncountable set is the real numbers between 0 and 1, denoted as the interval \[ \{ x \in \mathbb{R} : 0 < x < 1 \} \].These sets have more elements than any countable set and possess unique properties in measure theory.
Understanding whether a set is countable or noncountable is essential when applying concepts like outer measure, as they define whether the measure assigned is 0 or 1, respectively. This distinction in measure based on countability reflects the "size" or "cardinality" of the set in question, providing a foundational tool in mathematical analysis and probability.
Understanding whether a set is countable or noncountable is essential when applying concepts like outer measure, as they define whether the measure assigned is 0 or 1, respectively. This distinction in measure based on countability reflects the "size" or "cardinality" of the set in question, providing a foundational tool in mathematical analysis and probability.
Measurable Sets
In measure theory, understanding measurable sets is vital. A set is considered measurable if it satisfies certain conditions with respect to a given measure. Specifically, for our outer measure \( \mu^{*} \), a set \( E \) is defined as \( \mu^{*} \)-measurable if for every subset \( A \) of \( X \), the following equation holds:\[ \mu^{*}(A) = \mu^{*}(A \cap E) + \mu^{*}(A \setminus E) \].This condition ensures that the measure of \( A \) can be split perfectly into the measure of its intersection with \( E \) and the measure of the part not in \( E \).
Measurable sets are practical because they allow us to apply measures consistently across different parts of a set. In the given exercise, all subsets of \( X \) are found to be \( \mu^{*} \)-measurable, whether they are countable or noncountable. This property plays a significant role in simplifying many mathematical problems because it confirms that no subset of \( X \) is beyond the reach of this measure, ensuring consistency in analysis and integration.
Measurable sets are practical because they allow us to apply measures consistently across different parts of a set. In the given exercise, all subsets of \( X \) are found to be \( \mu^{*} \)-measurable, whether they are countable or noncountable. This property plays a significant role in simplifying many mathematical problems because it confirms that no subset of \( X \) is beyond the reach of this measure, ensuring consistency in analysis and integration.
Measure Theory
Measure theory is a branch of mathematical analysis focused on measuring sets in a rigorous way. The theory seeks to generalize the concept of length, area, and volume. In the context of the exercise, measure theory provides a framework to define and understand the ways we can "measure" noncountable sets.One of the key elements of measure theory is the concept of an outer measure, such as \( \mu^{*} \) in the problem. An outer measure provides a way to assign a size to every possible subset, including those that are highly irregular, by relying on three essential properties:
- \( \mu^{*}(\emptyset) = 0 \)
- \( \mu^{*}(A) \leq \mu^{*}(B) \) when \( A \subseteq B \)
- \( \mu^{*}(\bigcup_{i=1}^{\infty} E_i) \leq \sum_{i=1}^{\infty} \mu^{*}(E_i) \)