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Für jedes \(f, \in \mathcal{T}^{+}\)ist \(\mu_{f}: \mathfrak{A} \rightarrow \overline{\mathbb{R}}, \mu_{f}(A):=\int_{X} f \cdot \chi_{A} d \mu\) ein Maß auf \(\mathfrak{A}\).

Short Answer

Expert verified
\(\mu_{f}\) is a measure on \(\mathfrak{A}\).

Step by step solution

01

Non-negativity

To show non-negativity, we have to show that for any set \(A \in \mathfrak{A}\), we have \(\mu_{f}(A) \ge 0\). We have \(\mu_{f}(A) = \int_{X} f \cdot \chi_{A} d \mu\). The integrand \(f \cdot \chi_{A}\) is non-negative because \(f \in \mathcal{T}^{+}\) and \(\chi_{A}\) is a characteristic function. Since the integral of a non-negative function is also non-negative, we have \(\mu_{f}(A) \ge 0\). This shows non-negativity.
02

Null Empty Set

To show that a null set gives a zero measure, we need to show that for the empty set \(\emptyset\), we have \(\mu_{f}(\emptyset) = 0\). We have \(\mu_{f}(\emptyset) =\int_{X} f \cdot \chi_{\emptyset} d \mu\), where \(\chi_{\emptyset}\) is the characteristic function of the empty set, which is zero everywhere. As a result, \(f \cdot \chi_{\emptyset}\) is also zero everywhere and its integral (i.e., \(\mu_{f}(\emptyset)\)) is therefore also zero. Hence, \(\mu_{f}(A)\) satisfies the second property of a measure.
03

Countable Additivity

To verify countable additivity, for any countable collection \((A_i)_{i \in \mathbb{N}}\) of disjoint sets in \(\mathfrak{A}\), we have to show that \(\mu_{f}(\cup_{i=1}^\infty A_i) =\sum_{i=1}^\infty \mu_{f}(A_i)\). We have \(\mu_{f}(\cup_{i=1}^\infty A_i)\) = \(\int_{X} f \cdot \chi_{\cup_{i=1}^\infty A_i} d \mu\), and \(\sum_{i=1}^\infty \mu_{f}(A_i) =\sum_{i=1}^\infty \int_{X} f \cdot \chi_{A_i} d \mu\). With the countable additivity of the integral, the right-hand side equals the left-hand side, which verifies countable additivity. Hence, \(\mu_{f}(A)\) satisfies all three properties of a measure.

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Key Concepts

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

Non-negativity
When dealing with measure theory, a key concept is that of non-negativity. In measure theory, a measure assigns a non-negative number to every set within a certain collection of sets (called a sigma-algebra, here denoted as \(\mathfrak{A}\)). Non-negativity ensures that measures make sense in the real world where size or volume cannot be negative.
For example, consider the function \(\mu_f(A) = \int_X f \cdot \chi_A \, d\mu\). Here, \(f\) is a non-negative function because it's an element of \(\mathcal{T}^+\), ensuring that it doesn't take any negative values.
Additionally, \(\chi_A\) is a characteristic function for the set \(A\), and it only takes values of zero or one. Because both \(f\) and \(\chi_A\) are non-negative, the product \(f \cdot \chi_A\) is also non-negative.
  • This tells us that their integral, \(\mu_f(A)\), over a set \(A\) is always greater than or equal to zero.
  • Thus, non-negativity guarantees that the measure \(\mu_f\) aligns with our intuitive understanding of size or volume.
Null Empty Set
In measure theory, understanding the null condition for an empty set is important. The empty set, denoted as \(\emptyset\), inherently has no elements. This simplicity means its measure, or 'size', should naturally be zero.
For the measure \(\mu_f\), when applied to the empty set, it transforms to \(\mu_f(\emptyset) = \int_X f \cdot \chi_\emptyset \, d\mu\). Here, the characteristic function \(\chi_\emptyset\) of the empty set is zero at all points, because there are no points in \(\emptyset\).
  • This results in \(f \cdot \chi_\emptyset\) being zero everywhere, leading to an integral value of zero.
  • Thus, \(\mu_f(\emptyset) = 0\), satisfying the requirement of a null measure for the empty set.
Recognizing that the empty set has a measure of zero applies broadly across measure theory. It confirms that our mathematical framework is logically consistent with our expectation that no "size" exists for something with no elements.
Countable Additivity
Countable additivity, a cornerstone concept in measure theory, ensures that the measure of a infinite union of disjoint sets equals the sum of the measures of each set.
Consider a countable collection of disjoint sets \((A_i)_{i \in \mathbb{N}}\) in \(\mathfrak{A}\). For the measure \(\mu_f\), we want to confirm that
\[\mu_f\left(\bigcup_{i=1}^\infty A_i\right) = \sum_{i=1}^\infty \mu_f(A_i)\].
Using the definition, \(\mu_f\left(\bigcup_{i=1}^\infty A_i\right) = \int_X f \cdot \chi_{\bigcup_{i=1}^\infty A_i} \, d\mu\), and breaking it into parts gives
\[\sum_{i=1}^\infty \mu_f(A_i) = \sum_{i=1}^\infty \int_X f \cdot \chi_{A_i} \, d\mu\].
Since the integrals of sums of disjoint functions equal the sum of their integrals, we verify that
\[\int_X f \cdot \chi_{\bigcup_{i=1}^\infty A_i} \, d\mu = \sum_{i=1}^\infty \int_X f \cdot \chi_{A_i} \, d\mu\].
  • This principle ensures consistency in the way measures combine over infinite operations, crucial for building complex measures.
  • Countable additivity empowers us to extend measures beyond finite worlds and confirms that our measures uphold their properties over infinite scenarios.

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Most popular questions from this chapter

Es seien \(0

Ist \(f \in \mathcal{M}^{+}\)und \(\int_{X} f d \mu<\infty\), so gilt für jedes \(\varepsilon>0: \mu(\\{f>\varepsilon\\})<\infty\)

Erweitertes Lemma von Fatou: Die Funktionen \(f, f_{n}: X \rightarrow \overline{\mathbb{R}}\) seien meßbar und \(f\) quasiintegrierbar. a) Ist \(\int_{X} f d \mu>-\infty\) und \(f_{n} \geq f \mu\)-f.ü. \((n \in \mathrm{N})\), so gilt: $$ \int_{X} \varliminf_{n \rightarrow \infty} f_{n} d \mu \leq \varliminf_{n \rightarrow \infty} \int_{X} f_{n} d \mu $$ b) Ist \(\int_{X} f d \mu<\infty\) und \(f_{n} \leq f \mu\)-f.u. ( \(\left.n \in \mathbb{N}\right)\), so gilt: $$ \int_{X} \varlimsup_{n \rightarrow \infty} f_{n} d \mu \geq \varlimsup_{n \rightarrow \infty} \int_{X} f_{n} d \mu $$ c) Zeigen Sie, daß man oben auf die Voraussetzung \(\int_{X} f d \mu>-\infty\) bzw. \(\int_{X} f d \mu<\infty\) nicht verzichten kann und daß im Satz von der majorisierten Konvergenz die Bedingung der Existenz einer integrierbaren Majorante auch im Falle \(\mu(X)<\infty\) nicht durch die schwächere Bedingung \(\sup _{n \in \mathrm{N}} \int_{X}\left|f_{n}\right| d \mu<\infty\) ersetzt werden kann.

Es seien \(\left(a_{n}\right)_{n \geq 1}\) eine Folge positiver reeller Zahlen mit \(\sum_{n=1}^{\infty} a_{n} \log \left(1+1 / a_{n}\right)<\infty\) und \(\left(b_{n}\right)_{n \geq 1}\) eine beliebige Folge reeller Zahlen. Dann konvergiert die Reihe \(\sum_{n=1}^{\infty} a_{n} /\left|x-b_{n}\right| \lambda^{1}-\) f.ü. auf \(\mathbb{R}\). (Hinweise: Es gilt \(\lim _{n \rightarrow \infty} a_{n}=0\). Man setze \(f_{n}(x):=a_{n} /\left|x-b_{n}\right|\) für \(a_{n} \leq\left|x-b_{n}\right|\) und \(f_{n}(x)=0\) sonst. Mit \(A_{n}=\left\\{x: f_{n}(x) \neq a_{n} /\left|x-b_{n}\right|\right\\}\) gilt \(\sum_{n=1}^{\infty} \lambda^{1}\left(A_{n}\right)<\infty\) und \(\lambda^{1}\left(\varlimsup_{n \rightarrow \infty} A_{n}\right)=0 .\) Für jedes \(R>0\) gilt nun \(\int_{-R}^{R} \sum_{n=1}^{\infty} f_{n} d \lambda^{1}<\infty\), also konvergiert \(\sum_{n=1}^{\infty} f_{n}\) f.ü. auf \(\mathbb{R}\).)

Ist \(\mu\) das Zählmaß auf \(\mathfrak{P}(\mathrm{N})\), so ist eine Funktion \(f: \mathrm{N} \rightarrow \mathbb{K}\) genau dann integrierbar, wenn \(\sum_{n=1}^{\infty} f(n)\) absolut konvergiert, und dann gilt: \(\int_{\mathrm{N}} f d \mu=\sum_{n=1}^{\infty} f(n)\).

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