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add bonus in ana

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christian
2020-11-17 13:25:19 +01:00
parent 0558a13676
commit 52fcda66cb
2 changed files with 20 additions and 1 deletions
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ws2020/ana/uebungen/ana2.pdf
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ws2020/ana/uebungen/ana2.tex
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@@ -243,10 +243,29 @@
\[ \[
1 = \nu(E) \le \underbrace{\nu(E \cap A)}_{\le 1} + \underbrace{\nu(E \cap A^{c})}_{=0} 1 = \nu(E) \le \underbrace{\nu(E \cap A)}_{\le 1} + \underbrace{\nu(E \cap A^{c})}_{=0}
\implies \nu(E \cap A) = 1 \implies A \in \mathscr{M} \implies \nu(E \cap A) = 1 \implies A \in \mathscr{M}
.\] Also insgesamt $F \subseteq M$. .\] Also insgesamt $\mathscr{F} \subseteq \mathscr{M}$.
\end{itemize} \end{itemize}
\end{proof} \end{proof}
\end{enumerate} \end{enumerate}
\end{aufgabe} \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} \end{document}