Aufgabe 4.2 – Messbarkeit und Bildmaß

 

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.

  1. Zeige, dass f\quad \mathcal{B}-\mathcal{P}\left( \Omega \right)-messbar ist
  2. 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