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Chapter 5: Problem 1

Die \(\sigma\) - Algebra \(\mathcal{L}^{p} \otimes \mathfrak{L}^{q}\) ist in \(\mathfrak{L}^{p+q}\) echt enthalten. (Hinweis: Jeder Schnitt einer Menge aus \(\mathfrak{L}^{p} \otimes \mathfrak{L}^{q}\) ist Lebesgue-me脽bar.)

Short Answer

Expert verified
The sigma-algebra \(\mathcal{L}^{p} \otimes \mathfrak{L}^{q}\) is indeed properly contained in \(\mathfrak{L}^{p+q}\). This is shown by proving that while all sets in \(\mathcal{L}^{p} \otimes \mathfrak{L}^{q}\) are part of \(\mathfrak{L}^{p+q}\), there exists at least one set in \(\mathfrak{L}^{p+q}\) that is not in \(\mathcal{L}^{p} \otimes \mathfrak{L}^{q}\).

Step by step solution

01

Understand Notations and Entities

Sigma algebra \(\mathcal{L}^{p}\) and \(\mathfrak{L}^{q}\) are families of subsets of some set that include the empty set and are closed under countable unions and countable intersections. The tensor product of them \(\mathcal{L}^{p} \otimes \mathfrak{L}^{q}\) is the smallest sigma algebra containing all sets of the form \(A \times B\) with \(A \in \mathcal{L}^{p}\) and \(B \in \mathfrak{L}^{q}\). And \(\mathfrak{L}^{p+q}\) is the sigma algebra generated by combined powers of \(\mathcal{L}^{p}\) and \(\mathfrak{L}^{q}\).
02

Approach via Measurability

Next, as hinted by the problem, we use the fact that any section (or intersection) of a set from the sigma-algebra \(\mathcal{L}^{p} \otimes \mathfrak{L}^{q}\) is Lebesgue measurable. This essentially means that for every set belonging to the sigma-algebra \(\mathcal{L}^{p} \otimes \mathfrak{L}^{q}\), the measure of any countable intersection of this set is equal to the countable intersection of the measures. We will use this property to show the statement.
03

Show the Inclusion

Now, we need to show that every set in \(\mathcal{L}^{p} \otimes \mathfrak{L}^{q}\) is part of \(\mathfrak{L}^{p+q}\). This can be done by taking an element of \(\mathcal{L}^{p} \otimes \mathfrak{L}^{q}\), let's say \(C \in \mathcal{L}^{p} \otimes \mathfrak{L}^{q}\), and showing it's also in \(\mathfrak{L}^{p+q}\). Considering a countable intersection of the sets within \(\mathcal{L}^{p} \otimes \mathfrak{L}^{q}\) which is contained within \(\mathcal{C}\), it's straightforward to show that this property holds using the measurability of the set.
04

Show it's a Proper Inclusion

In the final step, we are to show that the inclusion is a proper one; that is, \(\mathcal{L}^{p} \otimes \mathfrak{L}^{q}\) is not the equal to \(\mathfrak{L}^{p+q}\). To do this we aim to find an example where some set, say \(D\), is in \(\mathfrak{L}^{p+q}\) but not in \(\mathcal{L}^{p} \otimes \mathfrak{L}^{q}\). Given the closure properties of \(\mathcal{L}^{p} \otimes \mathfrak{L}^{q}\), \(D\) could be represented as the union of certain sets not expressible as intersections within the set \(\mathcal{C}\) and thus fails to be contained within \(\mathcal{L}^{p} \otimes \mathfrak{L}^{q}\).

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Key Concepts

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

Sigma Algebra
A sigma algebra, denoted as \( \sigma \)-algebra, is a collection of sets that is closed under certain operations, forming the backbone of measure theory. It includes:
  • The empty set, ensuring every sigma algebra is never empty.
  • Closures under countable unions, meaning if you pick any collection of sets from the sigma algebra, their union is also part of this algebra.
  • Closures under countable intersections, similar to unions, but for intersections of sets.
  • Closures under complements, where the complement of any set in the sigma algebra is also within it.
These properties help in structuring measurable spaces where each "piece" or set can be measured accordingly. Understanding this foundation is crucial as it allows us to describe more complex structures like product and tensor product sigma algebras.
Lebesgue Measurable
Lebesgue measurable sets lie at the heart of measure theory. When we say a set is Lebesgue measurable, it means that it fits well within the framework of measure theory, allowing us to assign a measure (like length, area, or volume) to it.
Lebesgue measurability is essential because it helps in managing irregular or "dirty" data, making it usable for calculus and analysis. Properties of Lebesgue measurable sets include:
  • They can be decomposed into more straightforward pieces, usually intervals or union of intervals.
  • Operations like unions or intersections of these sets also yield Lebesgue measurable sets.
  • They interact smoothly with sigma algebras, making them indispensable for defining integrals and other analytical tools.
This concept allows for precise mathematical manipulation within the realm of functions and probabilities.
Tensor Product in Measure Theory
Tensor products in measure theory allow us to combine different sigma algebras into a new, larger sigma algebra. Consider a situation where we have two spaces each equipped with its own sigma algebra. The tensor product \( \mathcal{L}^{p} \otimes \mathfrak{L}^{q} \) does the following:
  • It forms the smallest sigma algebra containing all pairwise products \( A \times B \), where \( A \) is from the first space and \( B \) is from the second space.
  • This operation permits the handling of multi-dimensional data by extending the capacity of each original sigma algebra.
  • By ensuring closure properties similar to the individual algebras, it makes working with products of measurable spaces more coherent and reliable.
Understanding tensor products is essential as it extends the usefulness of sigma algebras beyond individual spaces to tackle complex multi-space problems.
Proper Inclusion in Sigma Algebras
A proper inclusion in sigma algebras indicates a situation where one sigma algebra is entirely contained within another, but they are not equal. In our context with \( \mathcal{L}^{p} \otimes \mathfrak{L}^{q} \) and \( \mathfrak{L}^{p+q} \), this means:
  • Every set in \( \mathcal{L}^{p} \otimes \mathfrak{L}^{q} \) is also found in \( \mathfrak{L}^{p+q} \).
  • However, there exist some sets in \( \mathfrak{L}^{p+q} \) that cannot be constructed within \( \mathcal{L}^{p} \otimes \mathfrak{L}^{q} \).
  • This difference arises due to the extra complexity or structure that \( \mathfrak{L}^{p+q} \) can accommodate vs the basic product structure in the tensor product.
Identifying proper inclusions is beneficial because it highlights set relationships' subtlety and provides insights into the hierarchy and capabilities of different sigma algebras.

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Most popular questions from this chapter

Es seien \(B \in \mathfrak{B}^{p}, \beta^{p}(B)<\infty\), und f眉r festes \(a \in \mathbb{R}^{p+1}\) mit \(a_{p+1}>0\) sei \(K\) der Kegel mit der Basis \(B\) und der Spitze \(a\), d.h. \(K=\\{\lambda(b, 0)+(1-\lambda) a: 0 \leq \lambda \leq 1, b \in B\\}\). Dann ist \(K \in \mathfrak{B}^{p+1}\) und $$ \beta^{p+1}(K)=\frac{a_{p+1}}{p+1} \beta^{p}(B). $$

In der Situation des Satzes von PLANCHEREL gilt f眉r alle \(x \in \mathbb{R}^{p}\) : $$ \int_{\mathbb{R}^{p}} f(t+x) \overline{f(t)} d \mu_{p}(t)=\int_{\mathbb{R}^{p}}|\hat{f}(t)|^{2} e^{i\langle t, x\rangle} d \mu_{p}(t) $$ (Hinweis: Grenz眉bergang \(n \rightarrow \infty\) in (3.8) oder Parsevalsche Formel.)

Ist \(X\) 眉berabz盲hlbar und \(\mathfrak{A}\) die von den endlichen Teilmengen von \(X\) erzeugte \(\sigma\)-Algebra 眉ber \(X\), so geh枚rt die Diagonale \(\triangle:=\\{(x, x): x \in X\\}\) nicht zu \(\mathfrak{A} \otimes \mathfrak{A}\), obwohl alle Schnitte von \(\triangle\) zu \(\mathfrak{A}\) geh枚ren. (Hinweis: Satz III.5.14.) F眉r die folgenden Aufgaben 1.4-1.6 gelten die Voraussetzungen und Bezeichnungen von Satz \(1.5 .\)

Pr眉fen Sie, welche der Integrale $$ \int_{0}^{1}\left(\int_{0}^{1} f(x, y) d x\right) d y, \int_{0}^{1}\left(\int_{0}^{1} f(x, y) d y\right) d x, \int_{I} f d \beta^{2} $$ \({ }^{9}\) Mit Blick auf heute bisweilen anzutreffende Verh盲ltnisse kann der Verf. ein \(鈥� \mathrm{O}\) tempora, o mores!" nicht unterdr眉cken. \((I=] 0,1\left[{ }^{2}\right)\) f眉r die folgenden Funktionen existieren und 眉bereinstimmen. a) \(f(x, y)=(x-y) /(x+y)^{3}\) f眉r \(x, y>0\). b) \(f(x, y)=\frac{\partial}{\partial x} \frac{\partial}{\partial y}\left(\left(x^{2}-y^{2}\right)^{2} /\left(x^{2}+y^{2}\right)^{2}\right)\) f眉r \(x, y>0\) c) \(f(x, y):= \begin{cases}2^{2 n} & \text { f眉r } 2^{-n}

Es seien \(0
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