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If \((X, \mathcal{M}, \mu)\) is a measure space and \(\left\\{E_{j}\right\\}_{1}^{\infty} \subset \mathcal{M}\), then \(\mu\left(\lim \inf E_{j}\right) \leq\) \(\liminf \mu\left(E_{j}\right) .\) Also, \(\mu\left(\lim \sup E_{j}\right) \geq \lim \sup \mu\left(E_{j}\right)\) provided that \(\mu\left(\bigcup_{1}^{\infty} E_{j}\right)<\) \(\infty\).

Short Answer

Expert verified
The inequalities \(\mu(\liminf E_j) \leq \liminf \mu(E_j)\) and \(\mu(\limsup E_j) \geq \limsup \mu(E_j)\) are shown by considering measure and set properties.

Step by step solution

01

Understanding the Terms

The limit inferior of a sequence of sets \(\{E_j\}\) is defined as \(\liminf E_j = \bigcup_{n=1}^{\infty} \bigcap_{j=n}^{\infty} E_j\). The limit superior is \(\limsup E_j = \bigcap_{n=1}^{\infty} \bigcup_{j=n}^{\infty} E_j\). These represent, respectively, the set of points that eventually belong to all but finitely many \(E_j\), and the set of points that belong to infinitely many \(E_j\).
02

Apply Measure Limit Inferior Inequality

To show \(\mu(\liminf E_j) \leq \liminf \mu(E_j)\), note that \(\bigcap_{j=n}^{\infty} E_j\) is decreasing in \(n\), hence \(\mu\left(\bigcap_{j=n}^{\infty} E_j\right)\) also forms a decreasing sequence. By continuity from above, \(\mu(\liminf E_j) = \lim_{n \to \infty} \mu\left(\bigcap_{j=n}^{\infty} E_j\right)\). Since \(\mu(E_j)\) is not necessarily non-increasing, \(\lim_{j\to\infty} \mu(E_j)\) defines the lim-inf of a sequence, ensuring \(\mu(\liminf E_j) \leq \liminf \mu(E_j)\).
03

Apply Measure Limit Superior Inequality

To show \(\mu(\limsup E_j) \geq \limsup \mu(E_j)\), observe that for the lim-sup, \(\bigcup_{j=n}^{\infty} E_j\) is increasing as \(n\) increases, implying \(\mu\left(\bigcup_{j=n}^{\infty} E_j\right) \leq \mu(\bigcup_{j=1}^{\infty} E_j) \lt \infty\). Thus \(\mu(\limsup E_j) = \mu\left(\bigcap_{n=1}^{\infty} \bigcup_{j=n}^{\infty} E_j\right) \geq \limsup \mu(E_j)\), utilizing the continuity from below and the assumption of finite measure.

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Key Concepts

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

Limit Inferior
The limit inferior, often denoted as \( \liminf \), of a sequence of sets \( \{E_j\} \) is a concept used to understand the behavior of sets as index \( j \) goes to infinity. The formula for this is given by:
  • \( \liminf E_j = \bigcup_{n=1}^{\infty} \bigcap_{j=n}^{\infty} E_j \)
This expression means that you first take the intersection of all sets from the \( n^{th} \) position onwards, and then take the union over all \( n \). Essentially, it is the set of elements that eventually belong to all but finitely many of the sets \( E_j \).
To understand this, think of it as the lasting behavior of elements across the sequence—they appear consistently in the tail end of the sequence of sets. Therefore, when applying measures in measure theory, the limit inferior can help identify elements that have a persistent presence and the corresponding measure gives the 'size' of such persistence.
Limit Superior
The limit superior, or \( \limsup \), is another concept to describe the asymptotic behavior of a sequence of sets \( \{E_j\} \). It provides a picture of the most inclusive behavior across the sequence. The formula is given by:
  • \( \limsup E_j = \bigcap_{n=1}^{\infty} \bigcup_{j=n}^{\infty} E_j \)
Here you take the union of sets starting from the \( n^{th} \) position and move onwards, followed by an intersection over all \( n \).
The result is the set of elements that appear in infinitely many of the \( E_j \) sets. This captures the ultimate presence of elements that appear repeatedly across the sequence, even if it is sporadic.
In measure terms, by analyzing \( \limsup \), you get an idea of the maximum potential 'size' of these elements' presence over all subsets. This helps in understanding the limit towards which the sequence of measures tends.
Continuity of Measure
Continuity of measure is a fundamental property in measure theory that deals with the behavior of measures under limits of sets and sequences of sets.
There are two key principles here:
  • **Continuity from Above**: If \( E_1 \supseteq E_2 \supseteq E_3 \cdots \) is a decreasing sequence of sets, then \( \mu(\bigcap_{j=1}^{\infty} E_j) = \lim_{j \to \infty} \mu(E_j) \).
  • **Continuity from Below**: If \( E_1 \subseteq E_2 \subseteq E_3 \cdots \) is an increasing sequence of sets, and \( \mu(\bigcup_{j=1}^{\infty} E_j) \lt \infty \), then \( \mu(\bigcup_{j=1}^{\infty} E_j) = \lim_{j \to \infty} \mu(E_j) \).
In context, continuity from above is used to assess \( \liminf \), ensuring that the measure of a sequence's persistent part doesn’t surpass the limit inferior of the measures.
Similarly, continuity from below supports the analysis of \( \limsup \), where finite measure limits ensure that repeated or sporadic presence doesn't exceed \( \limsup \) of measures. These principles ensure a rigorous understanding of how measures behave as the structure of sets changes.

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Most popular questions from this chapter

Let \(\mu\) be a finite measure on \((X, \mathcal{M})\), and let \(\mu^{*}\) be the outer measure induced by \(\mu\). Suppose that \(E \subset X\) satisfies \(\mu^{*}(E)=\mu^{*}(X)\) (but not that \(E \in \mathcal{M}\) ). a. If \(A, B \in \mathcal{M}\) and \(A \cap E=B \cap E\), then \(\mu(A)=\mu(B)\). b. Let \(\mathcal{M}_{E}=\\{A \cap E: A \in \mathcal{M}\\}\), and define the function \(\nu\) on \(\mathcal{M}_{E}\) defined by \(\nu(A \cap E)=\mu(A)\) (which makes sense by (a)). Then \(\mathcal{M}_{E}\) is a \(\sigma\)-algebra on \(E\) and \(\nu\) is a measure on \(\mathrm{M}_{E}\).

If \(\mathcal{M}\) is the \(\sigma\)-algebra generated by \(\mathcal{\varepsilon}\), then \(\mathcal{M}\) is the union of the \(\sigma\)-algebras generated by \(\mathcal{F}\) as \(\mathcal{F}\) ranges over all countable subsets of \(\mathcal{E}\). (Hint: Show that the latter object is a \(\sigma\)-algebra.)

There exists a Borel set \(A \subset[0,1]\) such that \(0

A family of sets \(\mathcal{R} \subset \mathcal{P}(X)\) is called a ring if it is closed under finite unions and differences (i.e., if \(E_{1}, \ldots, E_{n} \in \mathcal{R}\), then \(\bigcup_{1}^{n} E_{j} \in \mathcal{R}\), and if \(E, F \in \mathcal{R}\), then \(E \backslash F \in \mathcal{R}\) ). A ring that is closed under countable unions is called a \(\sigma\)-ring. a. Rings (resp. \(\sigma\)-rings) are closed under finite (resp. countable) intersections. b. If \(\mathcal{R}\) is a ring (resp. \(\sigma\)-ring), then \(\mathcal{R}\) is an algebra (resp. \(\sigma\)-algebra) iff \(X \in \mathcal{R}\). c. If \(\mathcal{R}\) is a \(\sigma\)-ring, then \(\left\\{E \subset X: E \in \mathcal{R}\right.\) or \(\left.E^{c} \in \mathcal{R}\right\\}\) is a \(\sigma\)-algebra. d. If \(\mathcal{R}\) is a \(\sigma\)-ring, then \(\\{E \subset X: E \cap F \in \mathcal{R}\) for all \(F \in \mathcal{R}\\}\) is a \(\sigma\)-algebra.

Suppose \(\left\\{\alpha_{j}\right\\}_{1}^{\infty} \subset(0,1)\). a. \(\prod_{1}^{\infty}\left(1-\alpha_{j}\right)>0\) iff \(\sum_{1}^{\infty} \alpha_{j}<\infty\). (Compare \(\sum_{1}^{\infty} \log \left(1-\alpha_{j}\right)\) to \(\left.\sum \alpha_{j} .\right)\) b. Given \(\beta \in(0,1)\), exhibit a sequence \(\left\\{\alpha_{j}\right\\}\) such that \(\prod_{1}^{\infty}\left(1-\alpha_{j}\right)=\beta\).

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