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Chapter 7: Problem 15

Ist \(\rho\) ein Ma脽 auf \(\mathfrak{A}\) und \(\cdot M \in \mathfrak{A}\), so sei \(\rho_{M}(A):=\rho(A \cap M) \quad(A \in \mathfrak{A}) .-\) Es seien nun \(\mu, \nu\) zwei \(\sigma\)-endliche MaBe auf 21 . Dann existiert eine Zerlegung \(X=S \cup E(S, E \in \mathfrak{A}, S \cap E=\emptyset\) ), so da脽 gilt: \(\mu=\mu_{S}+\mu_{E}, \nu=\nu_{S}+\nu_{E}, \nu_{S} \perp \mu_{S}, \nu_{E} \ll \mu_{E}\) und \(\mu_{E} \ll \nu_{E}\). Entsprechendes gilt f眉r \(\sigma\)-endliche signierte Ma脽e \(\mu, \nu\). (Hinweis: Nach dem Satz von RADON-NIKOD脻M gibt es \(f, g \in \mathcal{M}^{+}\), so da脽 \(\mu=f \odot \tau, \nu=g \odot \tau\), wobei \(\tau:=\mu+\nu\). Die Mengen \(S:=\\{f=0\\} \cup\\{g=0\\}\) und \(E:=S^{c}\) leisten das Verlangte.)

Short Answer

Expert verified
With the provided guidance, an appealing decomposition of the measure space X was found to be \(X=S \cup E\), where \(S\) is \(\{f=0\} \cup \{g=0\}\) and \(E\) is the complement of \(S\). Furthermore, the measures decompose to \(\mu=\mu_{S}+\mu_{E}\) and \(\nu=\nu_{S}+\nu_{E}\), such that \(\nu_{S} \perp \mu_{S}\), \(\nu_{E} \ll \mu_{E}\) and \(\mu_{E} \ll \nu_{E}\). The same properties apply to sigma-finite signed measures.

Step by step solution

01

Fetching the provided measures

From the text, given are two measures: \(\mu\) and \(\nu\), both of which are sigma-finite measures on 21, denoted \(\mathfrak{A}\). The problem also provides a hint using the Radon-Nikodym theorem. It specifies there exist two functions \(f, g \in \mathcal{M}^{+}\) such that \(\mu=f \odot \tau\) and \(\nu=g \odot \tau\), with \(\tau:=\mu+\nu\).
02

Decomposition of the space using Radon-Nikodym theorem

From the Radon-Nikodym theorem, it is known that the decomposition of space X is \(X=S \cup E(S, E \in \mathfrak{A}, S \cap E=\emptyset\)\). In this case, the set \(S\) is \(\{f=0\} \cup \{g=0\}\) and \(E:=S^{c}\) is the complement of S. The theorem also asserts \(\mu=\mu_{S}+\mu_{E}\) and \(\nu=\nu_{S}+\nu_{E}\), with \(\nu_{S} \perp \mu_{S}\), \(\nu_{E} \ll \mu_{E}\) and \(\mu_{E} \ll \nu_{E}\).
03

Applying properties to sigma-finite signed measures

The statement also extends to sigma-finite signed measures \(\mu\) and \(\nu\). The same decomposition and properties apply to these measures implying the theorem is valid for them.

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Key Concepts

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

Sigma-finite Measure
In measure theory, sigma-finite measures have a special role due to their manageable nature. A measure \( \mu \) is called sigma-finite if the entire space \( X \) can be divided into a countable union of measurable sets, each with finite measure. This decomposition allows for more straightforward analysis and integration over these sets.
Sigma-finite measures facilitate the use of the Radon-Nikodym theorem. This theorem requires the measures to be sigma-finite to ensure the existence of derivative-like functions between measures. The flexibility of sigma-finite measures lies in this ability to handle complex spaces by breaking them down into simpler, finite parts.
For example:
  • If \( X = \bigsqcup_{i=1}^{ \infty} X_i \)
  • and each \( \mu(X_i) < \infty \), then \( \mu \) is sigma-finite.
This concept allows us to unify the behavior of measures across different types of spaces, making advanced theorems applicable by ensuring analytic properties are present.
Measure Decomposition
Measure decomposition is a pivotal concept in the field of measure theory. It refers to the breaking down of a measure into distinct parts or components that are easier to analyze or understand. The Radon-Nikodym theorem provides a framework for such decomposition, offering a way to express one measure in terms of another, given they are absolutely continuous with respect to each other.
The exercise outlines a specific decomposition of the space \( X \), where \( X = S \cup E \) and \( S \cap E = \emptyset \). This means \( S \) and \( E \) are disjoint sets covering the entire space. The decomposition rests on properties such as mutual absolute continuity (\( u_E \ll \mu_E \) and \( \mu_E \ll u_E \)) and singularity (\( u_S \perp \mu_S \)).
By identifying functions \( f \) and \( g \) satisfying \( \mu = f \odot \tau \) and \( u = g \odot \tau \), with \( \tau = \mu + u \), the decomposition becomes evident:
  • Set \( S = \{ f = 0 \} \cup \{ g = 0 \} \)
  • Set \( E = S^c \)
These decompositions help in mapping out how different measures relate and interact within a given space, paving the way for further analysis or measure construction.
Signed Measures
Signed measures extend the concept of regular, non-negative measures by allowing them to take on negative values. This is similar to how signed numbers extend the realm of natural numbers.
In the context of the exercise, we consider sigma-finite signed measures \( \mu \) and \( u \). Even though they are not strictly positive, the principles from measure theory, such as the Radon-Nikodym theorem, can still be applied. This is crucial for problems requiring the integration of more complex or fluctuating quantities.
Characteristics of signed measures:
  • They can assume negative values, allowing them to represent more sophisticated phenomena.
  • They are composed of a positive part and a negative part.
  • When decomposed, the positive and negative parts aid in studying their attributes and implications.
The application of regular measure decomposition techniques to signed measures expands their usability, ensuring that standard measure-theoretic approaches can be adapted for broader classes of mathematical objects or spaces.

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Most popular questions from this chapter

Es seien \(\mu\) ein endliches signiertes \(\mathrm{Ma} B, \nu\) ein signierter Inhalt auf \(\mathfrak{A}\), und f眉r jede Folge \(\left(A_{n}\right)_{n \geq 1}\) von Mengen aus \(\mathfrak{A}\) mit \(\mathbf{|} \mu \mid\left(A_{n}\right) \rightarrow 0\) gelte \(\nu\left(A_{n}\right) \rightarrow 0\). Dann ist \(\nu\) ein signiertes \(\mathrm{MaB}\) auf \(\mathfrak{A}\).

Ist \((X, \mathfrak{A}, \mu)\) ein MaBraum und \(f: X \rightarrow \mathbb{C}\) integrierbar, so ist \(\nu: \mathfrak{A} \rightarrow \mathbb{C}\), $$ \nu(A):=\int_{A} f d \mu \quad(A \in \mathfrak{A}) $$ ein komplexes MaB; \(\nu\) hei脽t das komplexe Ma脽 mit der Dichte \(f\) bez. \(\nu\); Schreibweise: \(\nu=f \odot \mu\). Zeigen Sie: \(|\nu|=|f| \odot \mu\).

a) Ist \(\nu: \mathfrak{A} \rightarrow \mathbb{C}\) ein komplexes MaB, so gibt es eine me脽bare Funktion \(g: X \rightarrow \mathbb{C}\) mit \(|g|=1\), so da脽 \(\nu=g \odot \mid \nu \|\). (Bemerkung: In Analogie zur Polarkoordinatendarstellung komplexer Zahlen nennt man diese Darstellung die polare Zerlegung von \(\nu\).) b) F眉r alle \(f \in \mathcal{L}_{C}^{1}(\nu)\) gilt $$ \left|\int_{X} f d \nu\right| \leq \int_{X}|f| d|\nu| $$

Eine Menge \(M \in \mathfrak{A}\) hei脽t eine lokale \((\mu-)\) Nullmenge, wenn f眉r alle \(E \in \mathfrak{A}\) mit \(\mu(E)<\infty\) gilt \(\mu(M \cap E)=0\) a) Jede Nullmenge ist eine lokale Nullmenge, und jede lokale Nullmenge von \(\sigma\)-endlichem MaB ist eine Nullmenge, aber eine lokale Nullmenge braucht keine Nullmenge zu sein. b) Die Abbildung \(\varphi: L^{\infty} \rightarrow\left(L^{1}\right)^{\prime}\) aus Satz \(3.2\) ist injektiv genau dann, wenn jede lokale Nullmenge eine Nullmenge ist.

Ist \(N \subset \mathbb{R}\) eine \(\lambda\)-Nullmenge, so existiert eine wachsende absolut stetige Funktion \(f\) : \(\mathbb{R} \rightarrow \mathbb{R}\), die in keinem \(x \in N\) differenzierbar ist. (Hinweise: Es seien \(U_{n} \supset N, U_{n}\) offen, \(\lambda\left(U_{n}\right)<2^{-n} \quad(n \geq 1)\) und \(\left.f(x):=\int_{0}^{x} \sum_{n=1}^{\infty} \chi_{U_{n}}(t) d t \quad(x \in \mathbb{R}) .\right)\)
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