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@@ -243,10 +243,29 @@ |
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\[ |
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1 = \nu(E) \le \underbrace{\nu(E \cap A)}_{\le 1} + \underbrace{\nu(E \cap A^{c})}_{=0} |
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\implies \nu(E \cap A) = 1 \implies A \in \mathscr{M} |
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.\] Also insgesamt $F \subseteq M$. |
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.\] Also insgesamt $\mathscr{F} \subseteq \mathscr{M}$. |
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\end{itemize} |
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\end{proof} |
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\end{enumerate} |
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\end{aufgabe} |
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\begin{aufgabe} |
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Beh.: $\mu$ ist ein Maß. |
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\begin{proof} |
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\begin{enumerate}[(i)] |
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\item $\mu(\emptyset) = 0$, denn $\#(\emptyset \cap \{1, \ldots, n\}) = 0$ $\forall n \in \N$. |
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\item Sei $A_i \in \mathscr{P}(\N)$ für $i \in \N$ und $A_i \cap A_j = \emptyset$ für $i\neq j$. |
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Dann gilt |
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\begin{salign*} |
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\mu\left( \bigcupdot_{i \in \N} A_i \right) |
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&= \limsup_{n \to \infty} \frac{1}{n} \# \left( \bigcupdot_{i \in \N} A_i \cap \{1, \ldots, n\} \right) \\ |
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&= \limsup_{n \to \infty} \frac{1}{n} \# \left( \bigcupdot_{i \in \N} (A_i \cap \{1, \ldots, n\} ) \right) \\ |
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&\stackrel{\text{disj. Ver.}}{=} \limsup_{n \to \infty} \frac{1}{n} \sum_{i \in \N} \#(A_i \cap \{1, \ldots, n\}) \\ |
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&= \sum_{i \in \N} \limsup_{n \to \infty} \frac{1}{n} \#(A_i \cap \{1, \ldots, n\}) \\ |
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&= \sum_{i \in \N} \mu(A_i) |
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.\end{salign*} |
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\end{enumerate} |
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\end{proof} |
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\end{aufgabe} |
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\end{document} |
