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Author SHA1 Message Date
christian 0fbddc15c8 ignore xopp autosave files 2020-11-17 13:26:51 +01:00
christian addd2f84d9 add ana tut and theo sheet xopps 2020-11-17 13:26:39 +01:00
christian 52fcda66cb add bonus in ana 2020-11-17 13:25:19 +01:00
6 changed files with 21 additions and 1 deletions
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*.fdb_latexmk
*.toc
*.dat
*.autosave.xopp
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@@ -243,10 +243,29 @@
\[
1 = \nu(E) \le \underbrace{\nu(E \cap A)}_{\le 1} + \underbrace{\nu(E \cap A^{c})}_{=0}
\implies \nu(E \cap A) = 1 \implies A \in \mathscr{M}
.\] Also insgesamt $F \subseteq M$.
.\] Also insgesamt $\mathscr{F} \subseteq \mathscr{M}$.
\end{itemize}
\end{proof}
\end{enumerate}
\end{aufgabe}
\begin{aufgabe}
Beh.: $\mu$ ist ein Maß.
\begin{proof}
\begin{enumerate}[(i)]
\item $\mu(\emptyset) = 0$, denn $\#(\emptyset \cap \{1, \ldots, n\}) = 0$ $\forall n \in \N$.
\item Sei $A_i \in \mathscr{P}(\N)$ für $i \in \N$ und $A_i \cap A_j = \emptyset$ für $i\neq j$.
Dann gilt
\begin{salign*}
\mu\left( \bigcupdot_{i \in \N} A_i \right)
&= \limsup_{n \to \infty} \frac{1}{n} \# \left( \bigcupdot_{i \in \N} A_i \cap \{1, \ldots, n\} \right) \\
&= \limsup_{n \to \infty} \frac{1}{n} \# \left( \bigcupdot_{i \in \N} (A_i \cap \{1, \ldots, n\} ) \right) \\
&\stackrel{\text{disj. Ver.}}{=} \limsup_{n \to \infty} \frac{1}{n} \sum_{i \in \N} \#(A_i \cap \{1, \ldots, n\}) \\
&= \sum_{i \in \N} \limsup_{n \to \infty} \frac{1}{n} \#(A_i \cap \{1, \ldots, n\}) \\
&= \sum_{i \in \N} \mu(A_i)
.\end{salign*}
\end{enumerate}
\end{proof}
\end{aufgabe}
\end{document}
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