User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

Proposition 1.23
Let $\mathcal E$ be a set of sets which are subsets of some set $X$.

Let $\sigma\left({\mathcal E}\right)$ be the $\sigma$-algebra generated by $\mathcal E$.

Then $\sigma\left({\mathcal E}\right)$ can be constructed inductively.

The construction is as follows:

Let $\Omega$ denote the minimal uncountable well-ordered set.

Let $\alpha$ be an arbitrary initial segment in $\Omega$.

Considering separately the cases whether or not $\alpha$ has an immediate predecessor $\beta$, we define:


 * $\mathcal E_1 = \mathcal E$


 * $\mathcal E_\alpha = \begin{cases} \left \{ { \mathcal S \in \mathcal P\left({\mathcal E_\beta}\right) : \mathcal S \text { is countable or } \mathcal S^\complement \text{ is countable} }\right\} & \alpha \text{ has an immediate predecessor } \beta \\ \displaystyle \bigcup_{\beta \mathop \prec \alpha } \mathcal E_\beta & \text{ otherwise } \end{cases}$


 * $\mathcal E_{\Omega} = \displaystyle \bigcup_{\alpha \mathop \in \Omega} \mathcal E_{\alpha}$

Then $\sigma\left({\mathcal E}\right) = \mathcal E_{\Omega}$.

Step 1
We will show that $\sigma\left({\mathcal E}\right) \subseteq \mathcal E_{\Omega}$.

Define:


 * $\mathcal O = \left \{ { o \in \Omega: \mathcal E_o \text{ is } \sigma\left({\mathcal E}\right)}\text{-measurable} \right\}$

By the definition of a $\sigma$-algebra:


 * $\mathcal E_1 \subseteq \sigma\left({\mathcal E}\right)$.

From $\sigma$-Algebra of Countable Sets, if $\beta$ immediate predeces $\alpha$, then $\mathcal E_\alpha$ is a $\sigma$-algebra containing $\mathcal E_\beta$.

If $\beta$ strictly precedes $\alpha$ but is not an immediate predecessor of $\alpha$, then $\mathcal E_\alpha$ is a countable union of measurable sets.

By the definition of a $\sigma$-algebra and of union, $\mathcal E_\alpha$ is a $\sigma$-algebra containing $\mathcal E_\beta$.

Thus the hypotheses of well-ordered induction are satisfied, and $\mathcal O = \Omega$.

Thus $\mathcal E_\alpha \subseteq \sigma\left({\mathcal E}\right)$ for all $\alpha \in \Omega$.

By the properties of a $\sigma$-algebra:


 * $\displaystyle \bigcup_{\alpha \mathop \in \Omega} \mathcal E_\alpha \subseteq \sigma\left({\mathcal E}\right)$

as this is a countable union of measurable sets.

Thus $\mathcal E_{\Omega} \subseteq \sigma\left({\mathcal E}\right)$.

Step 2
By the definition of a $\sigma$-algebra generated by $\mathcal E$, the reverse inclusion:


 * $\mathcal E_{\Omega} \supseteq \sigma\left({\mathcal E}\right)$

will follow if $\mathcal E_{\Omega}$ is a $\sigma$-algebra.

Note that $\mathcal E = \mathcal E_0 \in E_{\Omega}$.

Let:


 * $\left\{ { E_1,E_2,E_3,\ldots } \right\}_{j \mathop \in \N}$

by an arbitrary countable indexed collection of sets in $\mathcal E_{\Omega}$.

By the inductive construction of $\mathcal E_{\Omega}$, for all $j \in \N$:


 * $E_j,E_j^{\complement} \in \mathcal{E}_{\alpha_j}$

for some initial segment $\alpha_j$.

Thus $\mathcal E_{\Omega}$ is closed under complement.

The set:


 * $\left \{ { \alpha_1, \alpha_2, \alpha_3, \ldots } \right\}$

is a countable subset of $\Omega$.

By Countable Subset of Minimal Uncountable Well-Ordered Set Has Upper Bound, it has an upper bound. Call it $\gamma$.

Then $E_j \in \mathcal E_{\gamma}$ for all $j \in \N$.

Hence for any $\delta$ strictly succeeding $\gamma$:


 * $\displaystyle \bigcup_{j \mathop \in \N} E_j \in \mathcal E_{\delta}$

Thus by the constructon of $\mathcal E_{\Omega}$:


 * $\mathcal E_{\delta} \subseteq \mathcal E_{\Omega}$

This means that $\mathcal E_{\Omega}$ is closed under countable union.

So $\mathcal E_{\Omega}$ is indeed a $\sigma$-algebra.

The result follows from the definition of set equality.

Corollary
Let $\mathcal E$ such that:


 * $\operatorname{card}\left({\N}\right) \le \operatorname{card}\left({\mathcal E}\right) \le \mathfrak c$

Then $\operatorname{card}\left({ \sigma\left({\mathcal E}\right) }\right) = \mathfrak c$.

Proof of Corollary
By the main result:


 * $\displaystyle \bigcup_{\alpha \mathop \in \Omega} \mathcal E_{\alpha} = \sigma\left({\mathcal E}\right)$

Thus by Leibniz's Law:


 * $\left \vert {\displaystyle \bigcup_{\alpha \mathop \in \Omega} \mathcal E_{\alpha} }\right \vert = \left \vert { \sigma\left({\mathcal E}\right) } \right \vert$

By the definition of union:


 * $\mathcal E_\alpha \subseteq \sigma\left({\mathcal E}\right)$ for all $\alpha \in \Omega$.

Thus:


 * $\left \vert { \mathcal E_\alpha }\right \vert \le \left \vert {\sigma\left({\mathcal E}\right)}\right \vert$ for all $\alpha \in \Omega$.

By Existence of Minimal Uncountable Well-Ordered Set:Corollary:


 * $\left \vert { \Omega } \right \vert \le \mathfrak c$

By Cardinality of Infinite Union of Infinite Sets:


 * $\left \vert {\displaystyle \bigcup_{\alpha \mathop \in \Omega} \mathcal E_{\alpha} }\right \vert \le \mathfrak c$

Thus:


 * $\left \vert { \sigma\left({\mathcal E}\right) } \right \vert \le \mathfrak c$

By Cardinality of Infinite Sigma-Algebra is at Least Cardinality of Continuum:


 * $\left \vert { \sigma\left({\mathcal E}\right) } \right \vert \ge \mathfrak c$

The result follows from the Cantor-Bernstein-Schröder Theorem.

Eventually
User:GFauxPas/Sandbox/Zeta2/lnxln1-x/existence

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/integrand

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/evaluation

User:GFauxPas/Sandbox/Zeta2/FourierSeries/

User:GFauxPas/Sandbox/Zeta2/Informal Proof

User:GFauxPas/Sandbox/NumberTheory