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If \(E \in \mathcal{L}\) and \(m(E)>0\), for any \(\alpha<1\) there is an open interval \(I\) such that \(m(E \cap I)>\alpha m(I)\).

Short Answer

Expert verified
Find an open interval \(I\) with \(m(E \cap I) > \alpha m(I)\) using properties of measure.

Step by step solution

01

Understanding the Problem

The problem asks us to find an open interval such that the measure of the intersection between this interval and a given set exceeds the measure of the interval scaled by a factor \(\alpha\), where \(\alpha < 1\).
02

Identifying the Set and Measure

We have a set \(E\) that's measurable and has a positive measure, \(m(E) > 0\). We need to find an open interval \(I\) such that the measure of the intersection \(E \cap I\) is greater than \(\alpha m(I)\).
03

Using Measure Theory

Given that \(E\) is measurable and \(m(E) > 0\), by the property of measures, for any \(\alpha < 1\), there exists an open interval \(I\) which intersects \(E\) such that \(m(E \cap I) > \alpha m(I)\). This is a consequence of the density of open sets in measure and the fact that \(E\) has positive measure.
04

Finding the Interval

Since \(m(E) > 0\), there exists an interval \(I\) with \(m(E \cap I) > 0\). Choose this interval small enough such that the measure of the part of \(I\) that lies in \(E\) is more than the fraction \(\alpha\) of the measure of \(I\), i.e., \(m(E \cap I) > \alpha m(I)\).
05

Conclusion

By carefully choosing the interval \(I\), we can always find an open interval such that \(m(E \cap I) > \alpha m(I)\) for any \(\alpha < 1\), leveraging the positive measure of \(E\) and properties of measurable sets.

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Key Concepts

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

Measurable Sets
Measurable sets are fundamental to measure theory, and understanding them is key for tackling many problems in this field. In simple terms, a measurable set is one for which a proper measure, often akin to "size" or "volume," can be assigned.
A classic example of such a measure is the length of an interval on a real number line. When a set is said to be measurable, it means we can use a measure (denoted as \(m\)) to quantify it in a meaningful way. Working within a measure space \((X, \mathcal{L}, m)\), any \(E\in\mathcal{L}\) indicates that the set \(E\) is measurable.
- **Importance**: Knowing a set is measurable ensures we can analyze and manipulate it using measure-theoretic methods.- **Applications**: Measurable sets are used in probability theory, where they help define events such that probabilities can be assigned.In our problem, the set \(E\) is measurable, and its measure \(m(E) > 0\). This guarantees the set's significance as it isn't negligible in size, meaning it necessarily occupies some "space." This property is crucial when looking to find intersections that also carry non-zero measure.
Intersection Measure
The concept of intersection measure comes into play when we want to find the measure of the overlap between two sets. Simply put, if two measurable sets \(A\) and \(B\) intersect, the measure \(m(A \cap B)\) will quantify the extent of their overlap.
In our exercise, we explore the measure of the intersection \(E \cap I\) where \(E\) is a measurable set and \(I\) is an open interval. The goal is to ensure \(m(E \cap I) > \alpha m(I)\) for some \(\alpha < 1\).
- **Properties**: - The measure \(m(E \cap I)\) must be positive if \(m(E)\) and \(m(I)\) both are positive. - Intersection measure is often less than or equal to the smallest measure of the intersecting sets.By using this concept, we ensure not only that the intersection is non-empty but also that its measure is sufficiently large relative to the measure of \(I\) scaled by \(\alpha\). This nuanced manipulation of intersections allows us to achieve specific problem requirements efficiently.
Open Interval
An open interval is a basic yet vital concept in real analysis and measure theory. An open interval \((a, b)\) on the real line is defined such that it includes all numbers between \(a\) and \(b\) but not \(a\) or \(b\) themselves.
In measure theory, open intervals are endowed with positive length, which serves as their measure. The positivity of this measure plays a fundamental role in various measure-theoretic applications.
- **Characteristics**: - Open intervals can be used to approximate more complex measurable sets due to their simple structure. - They are often employed in constructing other types of sets, thanks to their measure properties.In our context, the open interval \(I\) is integral, as it is employed to explore the intersection with \(E\). By adjusting \(I\)'s length and position suitably, we can control and optimize the intersection measure \(m(E \cap I)\), ensuring it surpasses \(\alpha m(I)\). Therefore, the open interval is not only a basic "building block" in measure theory but also a strategic tool to meet certain criteria in set operations.

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Most popular questions from this chapter

Let \(\mathcal{M}\) be an infinite \(\sigma\)-algebra. a. M contains an infinite sequence of disjoint sets. b. \(\operatorname{card}(\mathcal{M}) \geq c\).

Let \((X, \mathcal{M}, \mu)\) be a finite measure space. a. If \(E, F \in \mathcal{M}\) and \(\mu(E \Delta F)=0\), then \(\mu(E)=\mu(F)\). b. Say that \(E \sim F\) if \(\mu(E \Delta F)=0\); then \(\sim\) is an equivalence relation on \(\mathcal{M}\). c. For \(E, F \in \mathcal{M}\), define \(\rho(E, F)=\mu(E \Delta F)\). Then \(\rho(E, G) \leq \rho(E, F)+\) \(\rho(F, G)\), and hence \(\rho\) defines a metric on the space \(\mathcal{M} / \sim\) of equivalence classes.

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 \(\mathrm{M}_{E}\).

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.

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