Wahrscheinlichkeitstheorie Übung - Beispiel 10

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Beispiel 10 (a)

 \mathcal A \quad \sigma \text{ - Algebra} \Rightarrow \sigma(\mathcal A) = \mathcal A \,

Definition von  \sigma(.) \,

 \Sigma = \{ \mathcal N \subseteq P(\Omega) : \mathcal A \in \mathcal N, \mathcal N \, \, \sigma \mathrm{-Algebra} \} \,
 \sigma(\mathcal A ) = \bigcap_{\mathcal N \in \Sigma} \mathcal N \Rightarrow \mathcal A \supseteq \bigcap_{\mathcal N \in \Sigma} \mathcal N \, weil  \mathcal A \supseteq \mathcal N  \,
 \mathcal A \in \Sigma \, da  \mathcal A\, \sigma\, -Algebra

Beispiel 10 (b)

 \mathcal A, \mathcal B \sigma{ - Algebren}, \mathcal A \subseteq \mathcal B \Rightarrow \sigma(\mathcal A ) \subseteq \sigma(\mathcal B )\,

 \sigma(\mathcal A ) = \mathcal A, \sigma(\mathcal B ) = \mathcal B \Rightarrow \mathcal A  \subseteq \mathcal B \Rightarrow \sigma(\mathcal A) \subseteq \sigma(\mathcal B) \,

Beispiel 10 (c)

 \mathcal M \subseteq \mathcal M' \subseteq \sigma(\mathcal M) \Rightarrow \sigma(\mathcal M') = \sigma(\mathcal M) \,

 \mathcal M ' \subseteq \sigma(\mathcal M ) \, 

d.h.   \mathcal M ' \subseteq \sigma(\mathcal M) \in \Sigma' = \{\mathcal N ' \subseteq P(\Omega) : \mathcal M ' \subseteq \mathcal N', \mathcal N' \, \, \sigma \text{-Algebra} \} \,

 \Rightarrow \sigma(\mathcal M') \subseteq \sigma(\mathcal M)  \,

 \mathcal M \subseteq \mathcal M' \Rightarrow \sigma(\mathcal M) \subseteq \sigma(\mathcal M') \,
Beweis:

  \Sigma = \{\mathcal N \subseteq P(\Omega) : \mathcal M \subseteq \mathcal N, \mathcal N \, \, \sigma \text{-Algebra} \} \,

  \Sigma' = \{\mathcal N' \subseteq P(\Omega) : \mathcal M' \subseteq \mathcal N', \mathcal N' \, \, \sigma \text{-Algebra} \} \,

Wenn  \mathcal M  \subseteq \mathcal M' \,, dann gilt  \mathcal N' \in \Sigma' \Rightarrow \mathcal N' \in \Sigma \,, d.h.  \Sigma' \subseteq \Sigma \Rightarrow \bigcap_{\mathcal N' \in \Sigma'} \mathcal N' \supseteq \bigcap_{\mathcal N \in \Sigma} \mathcal N  \,, d.h.  \sigma(M) \supseteq \sigma(\mathcal M') \,