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