/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Es seien \((Y, \mathfrak{C}, \nu... [FREE SOLUTION] | 91影视

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Es seien \((Y, \mathfrak{C}, \nu)\) ein endlicher Ma脽raum, \((X, d)\) ein separabler (!) metrischer Raum und \(f, f_{n}: Y \rightarrow X \quad(n \in \mathrm{N})\) me脽bare Funktionen mit \(f_{n} \rightarrow f\) n.M. (s. Aufgabe VI.4.5). Ferner seien \(\mu:=f(\nu), \mu_{n}:=f_{n}(\nu)\) die zugeh枚rigen Bildma脽e. Dann gilt: \(\mu_{n} \stackrel{w}{\rightarrow} \mu\). Insbesondere gilt \(\mu_{n} \stackrel{w}{\rightarrow} \mu\), falls \(f_{n} \rightarrow f \nu\)-f.眉.

Short Answer

Expert verified
Yes, by applying the theorem of dominated convergence for finitely additive set functions, we establish that \(\mu_n\) converges weakly to \(\mu\), given that \(f_n\) converges almost everywhere to \(f\).

Step by step solution

01

Understanding weak convergence of measures

Define weak convergence of measures. If a sequence of (probability) measures \(\mu_n\) converges to \(\mu\) weakly, it means that for every bounded continuous function \(g\), the measure's integral with \(g\) as the integrand under \(\mu_n\) converges to the corresponding integral under \(\mu\) as \(n\) goes to infinity.
02

Proving \(\mu_n\) converges to \(\mu\)

Next, select an arbitrary bounded continuous function \(g\). Then, the weak convergence property can be written as follows: \[\lim_{n \rightarrow \infty} \int g \, d\mu_n = \int g \, d\mu\] Since \(g\) is continuous on a separable metric space, \(g \circ f_n\) and \(g \circ f\) are measurable. Because \(f_n\) converges almost everywhere to \(f\), we can apply Gottlieb's Dominated Convergence Theorem (for finitely additive set functions). Thus, \[\lim_{n \rightarrow \infty} \int (g \circ f_n) \, d\nu = \int (g \circ f) \, d\nu\] Remember that \(\mu_n\) and \(\mu\) originate from \(f_n(\nu)\) and \(f(\nu)\) respectively. Therefore, we can infer that \(\mu_n\) indeed converges weakly to \(\mu\).

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Key Concepts

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

Measurable Functions
In the context of measure theory, a function is considered measurable if it aligns with the structure of the measurable space it maps from. Specifically, for a function \( f: Y \rightarrow X \) to be measurable, the pre-image of every open set in \( X \) should be a set that belongs to the \( \sigma\)-algebra \( \mathfrak{C} \) of \( Y \). This property ensures that you can perform integration and measure-theoretic operations over \( f \).
Understanding measurability is pivotal in defining how functions behave under transformation and integration.
A measurable function allows us to capture more abstract concepts of limits and convergence. Here are some key points about measurable functions:
  • They serve as the foundation for defining integrals over complex spaces.
  • The composition of measurable functions is also measurable, ensuring consistent behavior through multiple transformations.
  • They facilitate the application of convergence theorems, such as the Dominated Convergence Theorem, by ensuring that transformations maintain measurability.
When dealing with sequences of functions \( f_n: Y \rightarrow X \), knowing they are measurable helps in understanding their convergence behavior towards a function \( f \), especially in separable metric spaces.
Separable Metric Space
A separable metric space is a type of metric space that contains a dense countable subset. This means that within such a space, you can find a countable set of points such that every point in the space is either in that countable set or can be approached by points from that set.
This characteristic makes separable metric spaces very useful in analysis, as they allow for the application of convergence theorems and the simplification of many otherwise complex proofs.
Some important properties of separable metric spaces include:
  • They potentially cover familiar spaces like \( \mathbb{R}^n \) when \( n \) is finite.
  • Such spaces facilitate the construction of continuous functions and the application of compactness arguments.
  • They simplify the handling of limits and continuity, essential for discussing the weak convergence of measures.
The fact that our space \((X, d)\) is separable is integral to certain convergence theorems (like the Dominated Convergence Theorem) being applicable, ensuring that functions defined over these spaces retain their measurable characteristics when limits are taken.
Dominated Convergence Theorem
The Dominated Convergence Theorem (DCT) is a powerful tool in measure theory and integration, allowing us to interchange limits and integrals under certain conditions.
This theorem states that if you have a sequence of measurable functions \( f_n \) converging almost everywhere to a function \( f \), and if there exists an integrable function \( g \) that 鈥渄ominates鈥 \( f_n \) (meaning \(|f_n(x)| \leq g(x)\) for nearly all \( x \), in some measure sense), then:
  • The integral of \( f_n \) converges to the integral of \( f \) as \( n \) approaches infinity.
  • You can safely swap the limit and the integral signs: \[ \lim_{n \to \infty} \int f_n \, du = \int f \, du \]
This theorem is particularly crucial when dealing with the weak convergence of measures as presented in the exercise.
Using the DCT, one can justify the step where the integral of the compositions \( g \circ f_n \) converges to the integral of \( g \circ f \).
This interchange of limits and integration helps in demonstrating that weak convergence, i.e., \( \mu_n \stackrel{w}{\rightarrow} \mu \), holds under the given circumstances, ensuring a bridge between analytical and measure-theoretic concepts.

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Most popular questions from this chapter

Bestimmen Sie ein Haar-Integral und ein Haar-Ma脽 f眉r die additive Gruppe \(\mathbb{R} \times \mathbb{R}_{d}\), wobei \(\mathbb{R}_{d}\) die mit der diskreten Topologie versehene Gruppe \((\mathbb{R},+)\) bezeichne.

\(X:=N \cup\\{\infty\\}\) trage folgende Topologie: Alle Teilmengen von \(N\) seien offen, und eine Menge \(A \subset X\) mit \(\infty \in A\) hei脽e genau dann offen, wenn $$ \varliminf_{n \rightarrow \infty} \frac{1}{n}|A \cap\\{1, \ldots, n\\}|=1 $$ Dann ist \(X\) ein normaler Hausdorff-Raum, in dem jede kompakte Menge endlich ist. Das Z盲hlma \(\beta\) auf \(X\) ist nicht lokal-endlich, also kein Borel-Ma脽, und das Z盲hlma脽 ist von innen, aber nicht von au脽en regul盲r.

Es seien \(\mu, \mu_{n}(n \in \mathbb{N})\) endliche Ma脽e auf der \(\sigma\)-Algebra \(\mathfrak{A}\) 眉ber der Menge \(X\). Dann sind folgende Aussagen 盲quivalent: a) F眉r alle \(A \in \mathfrak{A}\) gilt \(\lim _{n \rightarrow \infty} \mu_{n}(A)=\mu(A)\). b) F眉r alle \(f \in \mathfrak{L}^{\infty}(X, \mathfrak{A}, \mu)\) gilt $$ \lim _{n \rightarrow \infty} \int_{X} f d \mu_{n}=\int_{X} f d \mu $$

Es seien \(\mu, \nu: \mathfrak{A} \rightarrow[0, \infty]\) von innen regul盲re Ma脽e auf der \(\sigma\)-Algebra \(\mathfrak{A} \supset \mathfrak{B}\), und f眉r alle \(K \in \mathfrak{K} \operatorname{mit} \mu(K)=0\) sei auch \(\nu(K)=0 .\) Zeigen Sie: \(\nu \ll \mu\).

Ist \(\mu: \mathfrak{B} \rightarrow[0, \infty]\) ein \(\mathrm{Ma} \beta\), so ist \(\eta: \mathfrak{P}(X) \rightarrow[0, \infty]\), $$ \eta(A):=\inf \\{\mu(U): U \subset A, U \text { offen }\\} \quad(A \subset X) $$ ein von auben regul盲res 盲u脽eres \(\mathrm{MaB}\). Ist \(\mu\) endlich (oder moderat), so gilt \(\mathfrak{B} \subset \mathfrak{A}_{\eta}\) (= \(\sigma\)-Algebra der \(\eta\)-me脽baren Mengen) genau dann, wenn \(\mu\) von au脽en regul盲r ist, d.h. wenn \(\eta \mid \mathfrak{B}=\mu\) ist.

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