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Chapter 3: Problem 12

Es seien \((X, \mathfrak{A})\) ein Me脽raum und \(V\) ein Vektorraum von beschr盲nkten \(\mathfrak{A}\)-me脽baren Funktionen \(f: X \rightarrow \mathbb{R}\) mit folgenden Eigenschaften: (a) \(1 \in V\). (b) F眉r alle \(f, g \in V\) gilt \(\max (f, g) \in V\). (c) F眉r jede wachsende, gleichm盲脽ig beschr盲nkte Folge \(\left(f_{n}\right)_{n \geq 1}\) von Funktionen aus \(V\) gilt \(\lim _{n \rightarrow \infty} f_{n} \in V\) Dann ist \(\mathfrak{B}:=\left\\{A \subset X: \chi_{A} \in V\right\\} \subset \mathfrak{A}\) eine \(\sigma\)-Algebra, und \(V\) ist der Raum aller beschr盲nkten \(\mathfrak{B}\)-me脽baren reellwertigen Funktionen auf \(X\). (Hinweis: \(\chi_{\\{f>\alpha\\}}=\lim _{n \rightarrow \infty} \inf \left(1, n(f-\alpha)^{+}\right)\).)

Short Answer

Expert verified
\(\mathfrak{B}\) is a \(\sigma\)-algebra because it contains \(X\), is closed under complementation, and is closed under countable unions. The vector space \(V\) consists of all bounded \(\mathfrak{B}\)-measurable real-valued functions on \(X\) as \(V\) meets the given conditions, and any bounded \(\mathfrak{B}\)-measurable function can be expressed as the limit of a uniformly bounded increasing sequence from \(V\).

Step by step solution

01

Prove that \(\mathfrak{B}\) is a \(\sigma\)-algebra

To show that \(\mathfrak{B}\) is a \(\sigma\)-algebra, three properties need to be established. First, show that \(X \in \mathfrak{B}\). Since \(1 \in V\), and \(1\) is the indicator function for \(X\), hence \(X \in \mathfrak{B}\). Second, show that \(\mathfrak{B}\) is closed under complementation. Assume \(A \in \mathfrak{B}\). Then, since \(1 - \chi_{A} = \chi_{A^c}\) and \(1, \chi_{A} \in V\) by assumption, it follows that \(\chi_{A^c} \in V\) and hence \(A^c \in \mathfrak{B}\). Third, show that \(\mathfrak{B}\) is closed under countable unions. Assume \(A_i \in \mathfrak{B}\) for \(i \geq 1\). Then use the property \(b)\) and the hint to write \(\chi_{\bigcup_{i\geq 1} A_i}\) as a limit of a function sequence from \(V\) to show that \(A_i \in \mathfrak{B}\) for every \(i\), implies \(\bigcup_{i\geq 1} A_i \in \mathfrak{B}\).
02

Show that \(V\) is the space of all bounded \(\mathfrak{B}\)-measurable real-valued functions

Any element of \(V\) is \(\mathfrak{B}\)-measurable by definition because \(V\) is the vector space of bounded \(\mathfrak{A}\)-measurable functions and \(\mathfrak{B}\) is a subset of \(\mathfrak{A}\). For the reverse inclusion, let \(f\) be any bounded \(\mathfrak{B}\)-measurable function. Show that \(f\), in this case, is the limit of a uniformly bounded increasing sequence from \(V\) using the properties \(b)\) and \(c)\) to confirm that \(f \in V\).

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Key Concepts

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

Measurable functions
In measure theory, measurable functions play a crucial role in analyzing and understanding how functions interact with different types of sets within a given measure space. A measurable function is one that is compatible with the structure of its domain's sigma-algebra.To break this down:
  • Given a measure space \((X, \mathfrak{A})\), a function \(f: X \to \mathbb{R}\) is \(\mathfrak{A}\)-measurable if the preimage of a Borel set under \(f\) is in \(\mathfrak{A}\).
  • This property ensures that the function respects the measurable structure imposed by the sigma-algebra on \(X\).
  • In our exercise, \(V\) represents a vector space of bounded \(\mathfrak{A}\)-measurable functions, emphasizing the boundedness and measurability as key characteristics.
Measurable functions are essential in defining integrals, probabilities, and other concepts that depend on evaluating various function properties over measurable subsets.
Sigma-algebras
A sigma-algebra (\(\sigma\)-algebra) on a set \(X\) is a collection of subsets of \(X\) that is closed under several operations necessary for measure theory.
  • A sigma-algebra must include the empty set and the entire set \(X\).
  • It must be closed under complement operations. This means if a set \(A\) is in the sigma-algebra, then its complement \(A^c\) also belongs to it.
  • Finally, it must be closed under countable unions. If a sequence of sets \(A_i\) is in the sigma-algebra, then the union \(\bigcup_{i=1}^{\infty} A_i\) should also be in the sigma-algebra.
In our context, the problem involves proving that a certain collection of subsets forms a sigma-algebra, denoted by \(\mathfrak{B}\). We explore how certain functions, known as indicator functions, show subsets to be part of a sigma-algebra. Understanding these properties provides the basis for measurable functions on more complex sets, extending beyond basic measurable spaces.
Vector spaces
Vector spaces are fundamental constructs in both pure and applied mathematics, forming the backbone of linear algebra and numerous mathematical theories.In the context of measure theory, the vector space \(V\) consists of bounded \(\mathfrak{A}\)-measurable functions. Here's what this involves:
  • Closure under addition: If functions \(f\) and \(g\) are in \(V\), then their sum \(f + g\) must also be in \(V\).
  • Closure under scalar multiplication: If \(f\) is in \(V\) and \(c\) is a scalar, then the function \(cf\) must be in \(V\).
  • In the given exercise, properties such as these are utilized alongside convergence properties within the space \(V\), illustrating how \(V\) captures the behavior of measurable functions under mathematical operations.
By understanding vector spaces such as \(V\) in measure theory, one gains insight into how spaces of functions can be manipulated, extended, and explored, facilitating a deeper understanding of function behavior in complex measurable contexts.

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Most popular questions from this chapter

Es seien \((X, \mathfrak{A})\) ein Me脽raum und \(V\) ein Vektorraum von Funktionen \(f: X \rightarrow \mathbb{R}\) mit folgenden Eigenschaften: (a) Jeder Limes einer wachsenden Folge von Funktionen aus \(V\) geh枚rt zu \(V\). (b) Es gibt eine Algebra \(\mathfrak{C} \subset \mathfrak{A}\) mit \(\sigma(\mathfrak{C})=\mathfrak{A}\), so daB f眉r alle \(A \in \mathfrak{C}\) gilt \(\chi_{A} \in V\). Dann enth盲lt \(V\) alle me脽baren Funktionen \(f:(X, \mathfrak{A}) \rightarrow(\mathbb{R}, \mathfrak{B})\). Insbesondere ist der Vektorraum aller me脽baren reellwertigen Funktionen auf \(X\) der kleinste Vektorraum von Funktionen \(f: X \rightarrow \mathbb{R}\) mit den Eigenschaften (a), (b). (Hinweis: Satz I.6.2 oder Satz I.6.8.)

Es gibt ein translationsinvariantes Ma脽 \(\mu: \mathfrak{B}^{1} \rightarrow \overline{\mathbb{R}}\), welches nicht bewegungsinvariant ist (d.h. welches nicht invariant ist bez. der Spiegelung \(\sigma: \mathbb{R} \rightarrow \mathbb{R}, \sigma(x)=-x \quad(x \in \mathbb{R}))\). (Bemerkung: Nach Korollar 2.4, \(2.8\) ist jedes translationsinvariante Ma脽 \(\nu\) auf \(\mathfrak{B}^{1}\) mit \(\nu([0,1])<\infty\) bewegungsinvariant. - Hinweise: Konstruieren Sie eine Borel-Menge \(C \subset[0,1]\), so da脽 f眉r jede Folge \(\left(a_{n}\right)_{n \in \mathrm{N}}\) reeller Zahlen gilt \(\sigma(C) \not \subset \bigcup_{n \in \mathrm{N}}\left(C+a_{n}\right)\), und definieren Sie \(\mu(A):=0\), falls \(\mathrm{zu}\) \(A \in \mathfrak{B}^{1}\) eine Folge \(\left(a_{n}\right)_{n \in \mathrm{N}}\) reeller Zahlen existiert mit \(A \subset \bigcup_{n \in \mathrm{N}}\left(C+a_{n}\right)\), und \(\mu(A):=\infty\) anderenfalls. Die Menge \(C\) aller \(x \in[0,1]\), die eine Entwicklung zur Basis 4 haben, in der die Ziffer 2 nicht vorkommt, leistet das Verlangte.)

a) Ist \(G \subset \mathbb{R}^{p}\) eine additive Untergruppe des \(\mathbb{R}^{p}\) mit \(G \in \mathfrak{L}^{p}, \lambda^{p}(G)>0\), so gilt \(G=\mathbb{R}^{p}\). b) Nach a) ist jede Lebesgue-me脽bare additive Untergruppe \(G \subsetneq \mathbb{R}^{p}\) eine \(\lambda^{p}\)-Nullmenge. Eine solche Untergruppe kann durchaus gleichm盲chtig zu \(\mathbb{R}\) sein, wie das folgende Beispiel (Fall \(p=1\) ) lehrt: Es sei \(G\) die von den Zahlen \(\sum_{n=0}^{\infty} a_{n} 10^{-n !} \quad\left(a_{n} \in\\{0,1, \ldots, 9\\}\right.\) f眉r alle \(n \geq 0\) ) erzeugte additive Untergruppe von \(\mathbb{R}\). Dann ist \(G\) gleichm盲chtig zu \(\mathbb{R}, G\) ist eine \(\lambda^{1}\)-Nullmenge, und \(G\) ist von erster Bairescher Kategorie.

Es seien \(\left(X_{\iota}, \mathfrak{A}_{l}\right),\left(Y_{\iota}, \mathfrak{B}_{\iota}\right) \quad(\iota \in I)\) nicht-leere Me脽r盲ume, \(f_{\iota}: X_{\iota} \rightarrow Y_{\iota}\) Abbildungen. Die Funktion \(f: \prod_{l \in I} X_{\iota} \rightarrow \prod_{\iota \in I} Y_{\iota}, f\left(\left(x_{\iota}\right)_{\iota \in I}\right):=\left(f_{\iota}\left(x_{\iota}\right)\right)_{\iota \in I}\) ist genau dann \(\otimes_{\iota \in I} \mathfrak{A}_{\iota}-\bigotimes_{\iota \in I} \mathfrak{B}_{\iota}\)-me脽bar, wenn alle \(f_{\iota}:\left(X_{\iota}, \mathfrak{A}_{\iota}\right) \rightarrow\left(Y_{\iota}, \mathfrak{B}_{\iota}\right) \quad(\iota \in I)\) me脽bar sind.

Die Abbildung \(f: \mathbb{R} \rightarrow \mathbb{R}^{2}, f(x):=(x, 0)^{t} \quad(x \in \mathbb{R})\) ist \(\mathfrak{B}^{1}-\mathfrak{B}^{2}\)-me脽bar, aber nicht \(\mathfrak{L}^{1}-\mathfrak{L}^{2}\)-me脽bar.
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