Chapter 13: Problem 2
Let \(\mu\) be a locally finite measure. Show that \(\mu(K)<\infty\) for any compact set \(K\).
Short Answer
Expert verified
The measure \( \mu(K) \) is finite for any compact set \( K \) because \( K \) can be covered by finitely many sets with finite measure.
Step by step solution
01
Define Local Finiteness of Measure
A measure \( \mu \) is called locally finite if for every point \( x \) in the space and every neighborhood \( U \) of \( x \), we have \( \mu(U) < \infty \). This implies that for every compact set \( K \), there exists an open cover consisting of sets with finite measure.
02
Use Compactness of K
Since \( K \) is a compact set and \( \mu \) is locally finite, every point in \( K \) has a neighborhood with finite measure. By the definition of compactness, \( K \) can be covered by a finite subcollection of these neighborhoods.
03
Select Finite Open Cover
Let \( \{ U_1, U_2, ..., U_n \} \) be a finite subcollection of open sets covering \( K \) such that \( \mu(U_i) < \infty \) for each \( U_i \). Therefore, by local finiteness, we know each \( \mu(U_i) \) is finite.
04
Apply Subadditivity of Measure
Due to the subadditive property of measure, which states that for any collection of sets \( A_i \), \( \mu(\bigcup_{i} A_i) \leq \sum_{i} \mu(A_i) \), it follows that \( \mu(K) \leq \mu(U_1) + \mu(U_2) + ... + \mu(U_n) < \infty \), as each term is finite.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Locally Finite Measure
In measure theory, a measure \( \mu \) is termed **locally finite** when, for any point \( x \) in a given space and every neighborhood \( U \) around this point, the measure \( \mu(U) \) is finite. This means:
- \( \mu(U) < \infty \) for every neighborhood \( U \).
- The concept ensures that, although the whole space might be infinite, small local sections of it can be measured and they have finite measure.
Compact Set
A **compact set** is a fundamental concept in topology and analysis. Intuitively, a compact set can be thought of as being both closed and bounded, although this definition applies directly in Euclidean spaces.
- Formally, a set \( K \) is compact if every open cover of \( K \) has a finite subcover.
- This essentially means that, regardless of how a space is covered by open sets, there can always be a finite number of those sets that still cover the space completely.
Subadditivity of Measure
The concept of **subadditivity** is a vital characteristic of measures. A measure \( \mu \) is said to be subadditive if, for any collection of sets \( \{A_i\} \), the measure of the union of these sets does not exceed the sum of their measures:
- Mathematically, this is expressed as \( \mu\left(\bigcup_{i} A_i\right) \leq \sum_{i} \mu(A_i) \).
- This property ensures that while combining multiple measurable sets, the total measure does not disproportionately increase.