Merge branch 'master' of gitea:christian/uni

2020-11-24 10:24:56 +01:00
parent 4bd5670ddd 70c1818ff1
commit be6de1a7b3
2 changed files with 46 additions and 0 deletions
@@ -0,0 +1,46 @@
+\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}