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| \documentclass{../../../lecture} | |||||
| \usepackage[]{mathrsfs} | |||||
| \begin{document} | |||||
| \begin{aufgabe} | |||||
| Beh.: $f$ genau dann messbar, wenn $f^{-1}(\mathscr{A}) \subset \mathscr{E}$. | |||||
| \begin{proof} | |||||
| ,,$\implies$'': trivial, denn $f$ messbar $\implies$ $f^{-1}(\mathscr{F}) \subset \mathscr{E}$ und da $\mathscr{A} \subset \mathscr{F}$, folgt | |||||
| $f^{-1}(\mathscr{A}) \subset \mathscr{E}$. | |||||
| ,,$\impliedby$'': Sei also $f^{-1}(\mathscr{F}) \subset \mathscr{E}$. | |||||
| Also $f^{-1}(\mathscr{F}) = \{ f^{-1}(A) \mid A \in \mathscr{F} \} \subset \mathscr{E}$. | |||||
| \[ | |||||
| \mathscr{K} := \{ A \in \mathscr{F} \mid f^{-1}(A) \in \mathscr{E}\} | |||||
| .\] | |||||
| Z.z.: $\mathscr{K}$ $\sigma$-Algebra. | |||||
| \begin{enumerate}[(i)] | |||||
| \item $Y \in \mathscr{K}$, denn | |||||
| $f^{-1}(Y) = X \in \mathscr{E}$, da $\mathscr{E}$ | |||||
| $\sigma$-Algebra. | |||||
| \item Sei $A \in \mathscr{A}$. Dann ist | |||||
| $f^{-1}(A) \in \mathscr{E}$ und damit | |||||
| $f^{-1}(A^{c}) = f^{-1}(A)^{c} \in \mathscr{E}$, da | |||||
| $\mathscr{E}$ $\sigma$-Algebra. | |||||
| \item Seien $A_i \in \mathscr{K}$ für $i \in \N$. Dann | |||||
| ist $\forall i \in \N$: $f^{-1}(A_i) \in \mathscr{E}$. Damit | |||||
| folgt, da $\mathscr{E}$ $\sigma$-Algebra: | |||||
| \[ | |||||
| f^{-1}\left(\bigcup_{i \in \N} A_i \right) | |||||
| = \bigcup_{i \in \N} f^{-1}(A_i) \in \mathscr{E} | |||||
| .\] | |||||
| \end{enumerate} | |||||
| Nach Voraussetzung ist $\mathscr{A} \subset \mathscr{K}$. Es | |||||
| ist $\mathscr{K} \subset \mathscr{F}$ und | |||||
| $\mathscr{K}$ $\sigma$-Algebra, die $\mathscr{A}$ enthält, damit | |||||
| folgt $\mathscr{F} = \sigma(\mathscr{A}) \subset \mathscr{K}$, | |||||
| also insgesamt $\mathscr{K} = \mathscr{F}$. Also | |||||
| folgt $\forall A \in \mathscr{F}\colon f^{-1}(A) \in \mathscr{E}$, also | |||||
| $f^{-1}(\mathscr{F}) \subset \mathscr{E}$. | |||||
| \end{proof} | |||||
| \end{aufgabe} | |||||
| \end{document} | |||||