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$\mathscr M$ is a measure iff it is continuous from above.Why?

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Hi. If you show your progress or thoughts, you are much more likely to get help. As it stands, very few people will want to do everything for you. –  rschwieb Oct 25 '12 at 13:46

1 Answer 1

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So let $\mathscr M' = \bigcup_{\mathscr F \subseteq \mathscr E \text{ countable}} \sigma(\mathscr F)$, we will prove $\mathscr M' = \mathscr M$.

First note, that for every countable $\mathscr F \subseteq \mathscr E$ we have $\sigma( \mathscr F) \subseteq \sigma(\mathscr E) = \mathscr M$, hence also $\mathscr M' \subseteq \mathscr M$.

For the other inclusion we will prove that $\mathscr M'$ is a $\sigma$-algebra containing $\mathscr E$, as $\mathscr M$ is the smallest such, we will obtain $\mathscr M \subseteq \mathscr M'$.

  • For $E \in \mathscr E$, $\{E\} \subseteq \mathscr E$ is countable and $E \in \sigma\{E\} \subseteq \mathscr M'$. So $ \mathscr E \subseteq \mathscr M'$.

  • Let $ \mathscr F \subseteq \mathscr E$ be countable, then $\emptyset \in \sigma(\mathscr F) \subseteq \mathscr M'$.

  • Let $A \in \mathscr M'$ be given, then there is a countable $ \mathscr F \subseteq \mathscr E$ with $A \in \sigma( \mathscr F)$. But then $\complement A \in \sigma( \mathscr F)$, as the latter is a $\sigma$-algebra, and hence $\complement A \in \mathscr M'$.

  • Let $A_i \in \mathscr M'$ for $i \in \mathbb N$. For each $i$ there is a countable $\mathscr F_i \subseteq \mathscr E$ with $A_i \in \sigma( \mathscr F_i)$. Now set $\mathscr F := \bigcup_{i\in \mathbb N} \mathscr F_i$. $\mathscr F$ is, as a countable union of such, a countable subset of $\mathscr E$ and we have $A_i \in \sigma(\mathscr F_i) \subseteq \sigma(\mathscr F)$ for each $i$. As $\sigma(\mathscr F)$ is a $\sigma$-algebra, $\bigcup_{i \in \mathbb N} A_i \in \sigma(\mathscr F)\subseteq \mathscr M'$.

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thank you very much from this complete answer!thanks again! –  Somaye Oct 25 '12 at 13:41

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