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Chapter 2: Problem 8

Ist \(A \subset \mathbb{R}^{p}\) konvex, \(\stackrel{\circ}{A} \neq \emptyset\) und \(\lambda^{p}(A)<\infty\), so ist \(A\) beschr盲nkt (und damit Jordanme脽bar).

Short Answer

Expert verified
The set \( A \) is bounded as it has non-empty interior and finite measure, implying it is not spread out infinitely. Therefore, dominated by a ball in \( \mathbb{R}^{p} \). Based on the properties of a bounded and convex set, the condition that the boundary has zero measure is satisfied, which implies that the set is Jordan measurable. Therefore, the assertion that if \( A \subset \mathbb{R}^{p} \) is convex, \( \stackrel{\circ}{A} \neq \emptyset \) and \( \lambda^{p}(A) < \infty \), then \( A \) is bounded (and thus Jordan measurable) is proved.

Step by step solution

01

Understand the notations and terms

Firstly, you need to be familiar with the given notations and terms. In this context, \( A \subset \mathbb{R}^{p} \) means A is a subset of the p-dimensional Euclidean space. A convex set is a set where, for every pair of points within the set, every point on the straight line segment that joins the pair of points is also within the set. The interior of a set \( A \), denoted by \( \stackrel{\circ}{A} \), represents all points of \( A \) that does not belong to the boundary of \( A \). \( \lambda^{p}(A) \) stands for the Lebesgue measure of the set \( A \) in the \( p \)-dimensional Euclidean space, and \( \lambda^{p}(A)<\infty \) indicates that \( A \) has finite measure.
02

Connect the concepts and infer

From the definition of a convex set, it follows that if a set \( A \) has non-empty interior, then it has at least one point. The set is also supposed to have finite measure. In simpler terms, this suggests that the size of the set is finite. If a set has finite size and contains at least one point, it would mean that the set can't be too spread out, hence, it must be bounded. Being bounded implies that there exists a ball in \( \mathbb{R}^{p} \) that can contain the set \( A \).
03

Connect to Jordan measurability

A set in \( \mathbb{R}^{p} \) is Jordan measurable if and only if its boundary has Lebesgue measure zero. For bounded convex set, this condition is satisfied. Hence, the set is Jordan measurable.

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Key Concepts

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

Lebesgue Measure
) for sets that don't stretch out infinitely, indicating they are somehow contained or bounded within the space.
Jordan Measurability
. Essentially, the boundary needs to be negligible in the sense that its Lebesgue measure should be zero. This connects beautifully with convex sets, which tend to have well-behaved boundaries.

For instance, consider a circular disk in \( \mathbb{R}^{2} \)鈥攊ts boundary is the circumference, which is a one-dimensional curve with no area, and hence has a Lebesgue measure of zero. Therefore, the disk is Jordan measurable. Extending this to our exercise, a convex set \( A \) with a non-empty interior and a finite Lebesgue measure is expected to be bounded and, as a consequence of its boundedness, its boundary will also be well-behaved, making it Jordan measurable.
Bounded Sets
on the expansion of the set. Since such sets are contained, they are amenable to methods of analysis that rely on sets being of a manageable, finite size.

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Most popular questions from this chapter

Es seien \(F: \mathbb{R}^{p} \rightarrow \mathbb{R}\) eine \(p\)-mal stetig differenzierbare Funktion und \(f:=\frac{\partial}{\partial x_{1}} \cdot \ldots \cdot \frac{\partial}{\partial x_{p}} F\). a) F眉r \(a, b \in \mathbb{R}^{p}\) ist \(\triangle_{a}^{b} F=\int_{a_{1}}^{b_{1}} \cdot \ldots \cdot \int_{a_{p}}^{b_{p}} f\left(x_{1}, \ldots, x_{p}\right) d x_{1} \cdot \ldots d x_{p}(p\)-fach iteriertes Riemannsches Integral). b) Ist \(f \geq 0\), so ist \(F\) wachsend und \(\mu_{F}\) ein Pr盲ma脽.

Es seien \(X\) ein metrischer (oder topologischer) Raum, und \(\mu, \nu: \mathfrak{B}(X) \rightarrow \overline{\mathbb{R}}\) seien zwei Ma脽e. Zeigen Sie: Stimmen \(\mu\) und \(\nu\) auf allen offenen (bzw. abgeschlossenen) Teilmengen 眉berein und gibt es eine Folge \(\left(A_{n}\right)_{n \geq 1}\) offener (bzw. abgeschlossener) Teilmengen von \(X\) mit \(\mu\left(A_{n}\right)<\infty(n \in \mathrm{N}), X=\bigcup_{n=1}^{\infty} A_{n}\), so ist \(\mu=\nu .\) Ist \(X\) ein Hausdorff-Raum, so gilt diese Aussage sinngem盲脽 auch f眉r kompakte Mengen.

Es seien \(X\) ein metrischer (oder topologischer) Raum und \(\mu\) ein endlicher Inhalt auf dem Halbring \(\mathfrak{H}\) 眉ber \(X\). Dann hei脽t \(\mu\) von innen regul盲r, wenn zu jedem \(\varepsilon>0\) und jedem \(A \in \mathfrak{H}\) ein \(K \in \mathfrak{H}\) existiert, so da脽 gilt: \(\bar{K}\) ist kompakt, \(\bar{K} \subset A, \mu(A) \leq \mu(K)+\varepsilon .\) Zeigen Sie: Ist \(\mu\) von innen regul盲r, so ist \(\mu\) ein Pr盲ma脽. (Hinweis: Ist \(\mu\) von innen regul盲r, so auch die Fortsetzung \(\nu\) von \(\mu\) auf den von \(\mathfrak{H}\) erzeugten Ring. \(\nu\) gen眉gt der Bedingung d) aus Satz 1.10.)

Es seien \(F: \mathbb{R} \rightarrow \mathbb{R}\) die Cantorsche Funktion und \(x, y \in C, x

Es seien \(\eta^{p}\) das 盲u脽ere Lebesguesche Ma脽, \(A \subset \mathbb{R}^{p}\), und es gebe ein \(\left.\alpha \in\right] 0,1[\), so da \(\beta\) f眉r alle \(I \in \mathfrak{I}^{p}\) gilt \(\eta^{p}(A \cap I) \leq \alpha \lambda^{p}(I) .\) Dann ist \(A\) eine Lebesguesche Nullmenge.
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