Chapter 7: Problem 4
Let \(\left\\{\mu_{n}\right\\}\) be measures on the measurable space \(\llbracket X, \delta \rrbracket\), let \(\mu_{n} \geqslant \mu_{n+1}\) for each \(n\), and let \(\mu_{1}(X)<\infty .\) Write \(\lim \mu_{n}(E)=\mu(E)\) for each \(E \in \mathcal{S}\). Show that \(\mu\) is a measure on \(\mathcal{S}\).
Short Answer
Expert verified
\(\mu\) is a measure because it is non-negative, assigns zero to the null set, and is countably additive.
Step by step solution
01
Verify Decreasing Sequence Condition
Given that \(\mu_{n} \geqslant \mu_{n+1}\) for each \(n\), the sequence \(\{\mu_{n}(E)\}\) is non-increasing for every set \(E\in \mathcal{S}\). This means \(\mu(E) = \lim_{n \to \infty} \mu_n(E)\) will exist for each \(E\), as it converges to some value due to the boundedness by \(\mu_1(X) < \infty\).
02
Check Measure Properties - Non-negativity
Since each \(\mu_n\) is a measure, \(\mu_n(E) \geq 0\) for all \(n\) and for all \(E \in \mathcal{S}\). By the limit of non-negative terms, \(\mu(E) = \lim_{n \to \infty} \mu_n(E) \geq 0\) for all \(E\). Thus, \(\mu\) is non-negative.
03
Validate Null Set Property
Consider any null set \(\emptyset \in \mathcal{S}\). For each \(n\), \(\mu_n(\emptyset) = 0\). Hence, \(\mu(\emptyset) = \lim_{n \to \infty} \mu_n(\emptyset) = 0\). This shows \(\mu\) assigns zero measure to the null set.
04
Demonstrate Countable Additivity
Take a countable collection of disjoint sets \(\{E_i\}_{i=1}^{\infty} \subseteq \mathcal{S}\). Since each \(\mu_n\) is a measure, \(\mu_n(\bigcup_{i=1}^{\infty} E_i) = \sum_{i=1}^{\infty} \mu_n(E_i)\). By taking the limit of both sides as \(n \to \infty\) and using the Monotone Convergence Theorem, \(\mu(\bigcup_{i=1}^{\infty} E_i) = \lim_{n \to \infty} \mu_n(\bigcup_{i=1}^{\infty} E_i) = \sum_{i=1}^{\infty} \lim_{n \to \infty} \mu_n(E_i) = \sum_{i=1}^{\infty} \mu(E_i)\). Therefore, \(\mu\) satisfies countable additivity.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Decreasing Sequence of Measures
In measure theory, the concept of a decreasing sequence of measures arises when considering a set of measures that steadily reduces. This is formalized as a sequence \(\{\mu_n\}\) where each measure \(\mu_n\) is greater than or equal to the next, \(\mu_{n+1}\), for all \(n\). This means, for each measurable set \(E\), the values \(\mu_n(E)\) form a non-increasing sequence.
Such a sequence is particularly useful because it ensures the existence of a limit for \(\mu(E) = \lim_{n \to \infty} \mu_n(E)\) as \(n\) becomes very large. Due to the property that \(\mu_1(X) < \infty\), the sequence \(\mu_n(E)\) doesn't drop to negative infinity, and hence converges to a finite value.
Understanding decreasing sequences is pivotal as it provides a framework to define new measures by taking limits of existing non-increasing ones.
Such a sequence is particularly useful because it ensures the existence of a limit for \(\mu(E) = \lim_{n \to \infty} \mu_n(E)\) as \(n\) becomes very large. Due to the property that \(\mu_1(X) < \infty\), the sequence \(\mu_n(E)\) doesn't drop to negative infinity, and hence converges to a finite value.
Understanding decreasing sequences is pivotal as it provides a framework to define new measures by taking limits of existing non-increasing ones.
Non-negativity in Measure
Every measure, \(\mu_n\), in a sequence is equipped with the property of being non-negative, meaning for any measurable set \(E\), \(\mu_n(E) \geq 0\). This derives from the basic axioms of measure theory, ensuring that measures can be thought of as generalized notions of 'size' or 'volume', which are inherently non-negative.
When you take the limit of a sequence of non-negative numbers, the resulting limit will also be non-negative. Hence, for our limit measure \(\mu(E) = \lim_{n \to \infty} \mu_n(E)\), non-negativity is preserved, with \(\mu(E) \geq 0\) for all measurable sets \(E\).
The preservation of non-negativity when transitioning from \(\mu_n\) measures to the limit measure \(\mu\) underpins a key property that all resulting measures must satisfy.
When you take the limit of a sequence of non-negative numbers, the resulting limit will also be non-negative. Hence, for our limit measure \(\mu(E) = \lim_{n \to \infty} \mu_n(E)\), non-negativity is preserved, with \(\mu(E) \geq 0\) for all measurable sets \(E\).
The preservation of non-negativity when transitioning from \(\mu_n\) measures to the limit measure \(\mu\) underpins a key property that all resulting measures must satisfy.
Null Set Property
The null set property is a straightforward yet crucial characteristic in measure theory. It asserts that any measure assigns a zero value to the empty set. For our sequence of measures \(\mu_n\), this property means \(\mu_n(\emptyset) = 0\) for every \(n\).
When considering the limit measure \(\mu\), constructed from \(\mu_n\), this null set property continues to hold. By taking the limit, we find \(\mu(\emptyset) = \lim_{n \to \infty} \mu_n(\emptyset) = 0\).
Ensuring this property holds is vital as it maintains consistency in the framework of measure theory; no measure can attribute anything greater than zero size to a non-existent set.
When considering the limit measure \(\mu\), constructed from \(\mu_n\), this null set property continues to hold. By taking the limit, we find \(\mu(\emptyset) = \lim_{n \to \infty} \mu_n(\emptyset) = 0\).
Ensuring this property holds is vital as it maintains consistency in the framework of measure theory; no measure can attribute anything greater than zero size to a non-existent set.
Countable Additivity
Countable additivity is one of the core axioms of measures. It tells us that for a countable collection of disjoint measurable sets, the measure of their union is the sum of their individual measures.
In a formal setting, if \(\{E_i\}_{i=1}^{\infty}\) is a collection of disjoint sets in \(\mathcal{S}\), then for any measure \(\mu_n\), we have \(\mu_n(\bigcup_{i=1}^{\infty} E_i) = \sum_{i=1}^{\infty} \mu_n(E_i)\).
This property follows naturally to the limit measure \(\mu\) as well. Through techniques such as the Monotone Convergence Theorem, we can prove that \(\mu(\bigcup_{i=1}^{\infty} E_i) = \sum_{i=1}^{\infty} \mu(E_i)\), thus maintaining countable additivity.
This characteristic ensures that adding up "pieces" of a measurable set to find the total measure is always consistent, which is foundational to the measure's ability to assess 'size' appropriately.
In a formal setting, if \(\{E_i\}_{i=1}^{\infty}\) is a collection of disjoint sets in \(\mathcal{S}\), then for any measure \(\mu_n\), we have \(\mu_n(\bigcup_{i=1}^{\infty} E_i) = \sum_{i=1}^{\infty} \mu_n(E_i)\).
This property follows naturally to the limit measure \(\mu\) as well. Through techniques such as the Monotone Convergence Theorem, we can prove that \(\mu(\bigcup_{i=1}^{\infty} E_i) = \sum_{i=1}^{\infty} \mu(E_i)\), thus maintaining countable additivity.
This characteristic ensures that adding up "pieces" of a measurable set to find the total measure is always consistent, which is foundational to the measure's ability to assess 'size' appropriately.