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For any family $B \subset 2^{X} $ there exists the smallest algebra A containing B, we call this algebra the algebra generated by the family B and denote it $\sum (B)$.

For any family $B \subset 2^{X} $ there exists the smallest sigma algebra A containing B, we call this sigma algebra the sigma algebra generated by the family B and denote it $\sigma(B)$.

Prove (for any family of sets $B$):

1) $\sum(\sum(B)) = \sum(B)$

2) $ \sigma(\sigma(B))=\sigma(B)$

3) $\sum(\sigma(B))=\sigma(\sum(B))=\sigma(B)$

Attempt of proof:

To be honest I can see these properties only in examples, but I don't know how to prove it in generally.

Example: $X=\{1,2,3 \}$, and considering $\sigma(\{1\})$ and $\sum(\{1\})$

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  • $\begingroup$ Are you sure about the definition? If you, you can deduce 1) and 2) by definition since $\Sigma(B)$ is already an algebra and $\sigma(B)$ is already a sigma algebra. $\endgroup$ – Mundron Schmidt Dec 18 '17 at 12:32
  • $\begingroup$ These definition are from my notebook. Also I know definitions with interesections. 1. the algebra generated by $B$ is the intersection of all algebras containing $B$ 2. the sigma algebra generated by $B$ is intersection of all sigma algebras containing $B$. $\endgroup$ – Hikicianka Dec 18 '17 at 12:47
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Proof of 1).

$\sum (\sum (B))$ is an algebra by definition. $\sum (\sum (B)) \supseteq \sum(B) \supseteq B$ again by definition.

It remains to prove that $\sum (\sum (B))$ is the smallest algebra containing $B$ (hence we will have the equality $\sum (\sum (B))= \sum (B)$).

Now, let $A$ be any algebra containing $B$, we need to show that $\sum (\sum (B)) \subseteq A$. Since $\sum(B)$ is the smallest algebra containing $B$, we have $\sum(B) \subseteq A$. Since $\sum (\sum (B))$ is the smallest algebra containing $\sum(B)$, we have $\sum (\sum (B)) \subseteq A$. This completes the proof of 1).

Proof of 2): the same argument as 1).

Proof of 3): One equality is proved by monotonicity (in the sense that $X \subseteq Y \Rightarrow \sigma (X) \subseteq \sigma (Y)$) $$\sigma ( \sigma (B)) = \sigma(B) \subseteq \sigma ( \Sigma (B)) \subseteq \sigma ( \sigma (B))$$ hence all these are equal.

Finally, recalling that all sigma-algebras are algebras (hence $\Sigma (X) \subseteq \sigma (X)$), $$\sigma (B) \subseteq \Sigma ( \sigma (B)) \subseteq \sigma ( \sigma (B))= \sigma (B)$$ hence all these are equal.

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1)

$\Sigma(B)$ is evidently the smallest algebra that contains $\Sigma(B)$ so $\Sigma(\Sigma(B))=\Sigma(B)$

2)

same reasoning as under 1) but now for $\sigma$-algebras.

3)

$\sigma(B)$ is evidently the smallest algebra that contains $\sigma(B)$ so $\Sigma(\sigma(B))=\sigma(B)$.

$\sigma(B)$ is an algebra that contains $B$ so that $B\subseteq\Sigma(B)\subseteq\sigma(B)$.

This implies that $\sigma(B)\subseteq\sigma(\Sigma(B))\subseteq\sigma(\sigma(B))=\sigma(B)$ hence $\sigma(\Sigma(B))=\sigma(B)$.

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