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)

Theorem
Let $\mathcal E$ such that:


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

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

Proposition 1.23
Let $\mathcal E \subseteq \mathcal P(E)$ be a set of subsets of $E$.

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 set of countable ordinals.

Let $\alpha, \beta$ be arbitrary initial segments in $\Omega$.

Considering separately whether or not $\beta$ immediately precedes $\alpha$, we define:


 * $\mathcal E_0 = \mathcal E$


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


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

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

Proof of Proposition 1.23
By the definition of a generated sigma algebra:


 * $\mathcal E_0 \subseteq \sigma\left({\mathcal E}\right)$.

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

The construction of $\left \langle { \mathcal E_{\alpha} } \right \rangle_{\alpha \mathop \in \Omega}$ is such that the hypotheses of induction on $\Omega$ are satisfied.

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.

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