Aufgabe 4.2 – Messbarkeit und Bildmaß – Mathematical Engineering

Gegeben sei der Maßraum $\left( {\mathbb{R},\mathcal{B},\lambda } \right)$ und der Messraum $\left( {\Omega ,\mathcal{P}\left( \Omega \right)} \right),\quad \Omega = \left\{ {-1,0,1} \right\}$ . Sei weiter

$f:\mathbb{R} \to \Omega ,\quad \omega \mapsto \left\{ {\begin{array}{*{20}{c}} {-1} & {falls} & {\omega \in \left] {0,5} \right]} \\ 0 & {falls} & {\omega = 0} \\ 1 & {falls} & {sonst} \\ \end{array} } \right.$

eine Funktion.

Zeige, dass $f\quad \mathcal{B}-\mathcal{P}\left( \Omega \right)$ -messbar ist
Bestimme das Bildmaß

Lösung

$f:\mathbb{R} \to \Omega = \left\{ {-1,0,1} \right\}$

$\omega \mapsto \left\{ {\begin{array}{*{20}{c}} - 1 & \forall & {\omega \in \left] {0,5} \right]} \\ 0 & {falls} & {\omega = 0} \\ 1 & {sonst} & {} \\ \end{array} } \right.$

a )

$\mathcal{P}\left( \Omega \right) = \left\{ {\emptyset ,\Omega ,\left\{ {-1} \right\},\left\{ 0 \right\},\left\{ 1 \right\},\left\{ {-1,0} \right\},\left\{ {-1,1} \right\},\left\{ {0,1} \right\}} \right\}$

Die Urbilder sind

$f^{-1} \left( \emptyset \right) = \emptyset \in \mathcal{B}$

$f^{-1} \left( \Omega \right) = \mathbb{R} \in \mathcal{B}$

$f^{-1} \left( {\left\{ {-1} \right\}} \right) = \left] {0,5} \right] \in \mathcal{B}$

$f^{-1} \left( {\left\{ 0 \right\}} \right) = \left\{ 0 \right\} \in \mathcal{B}$

$f^{-1} \left( {\left\{ 1 \right\}} \right) = \mathbb{R}{{\backslash }}\left[ {0,5} \right] \in \mathcal{B}$

$f^{-1} \left( {\left\{ {-1,0} \right\}} \right) = \left[ {0,5} \right] \in \mathcal{B}$

$f^{-1} \left( {\left\{ {-1,1} \right\}} \right) = \mathbb{R}{{\backslash }}\left\{ 0 \right\} \in \mathcal{B}$

$f^{-1} \left( {\left\{ {0,1} \right\}} \right) = \mathbb{R}{{\backslash }}\left] {0,5} \right] \in \mathcal{B}$

Daraus folgt:

$f\,\,ist\,\,\mathcal{B}-\mathcal{P}\left( \Omega \right)-messbar$

b )

$\mu \left( B \right) = \lambda \left( {f^{-1} \left( B \right)} \right)$

$\mu \left( \emptyset \right) = \lambda \left( {f^{-1} \left( \emptyset \right)} \right) = \lambda \left( \emptyset \right) = 0$

$\mu \left( \Omega \right) = \lambda \left( {f^{-1} \left( \Omega \right)} \right) = \lambda \left( \mathbb{R} \right) = \infty$

$\mu \left( {\left\{ {-1} \right\}} \right) = \lambda \left( {\left] {0,5} \right]} \right) = 5-0 = 5$

$\mu \left( {\left\{ 0 \right\}} \right) = \lambda \left( {\left\{ 0 \right\}} \right) = 0$

$\mu \left( {\left\{ 1 \right\}} \right) = \lambda \left( {\mathbb{R}{{\backslash }}\left[ {0,5} \right]} \right) = \infty$

$\mu \left( {\left\{ {-1,0} \right\}} \right) = \lambda \left( {\left[ {0,5} \right]} \right) = 5-0 = 5$

$\mu \left( {\left\{ {-1,1} \right\}} \right) = \lambda \left( {\mathbb{R}{{\backslash }}\left\{ 0 \right\}} \right) = \infty$

$\mu \left( {\left\{ {0,1} \right\}} \right) = \lambda \left( {\mathbb{R}{{\backslash }}\left] {0,5} \right]} \right) = \infty$

Lösung

a )

b )

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