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 set of countable ordinals.

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

Proof of Proposition 1.23
Define:


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

where $\Omega$ is the set of countable ordinals.

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

For the other inclusion, consider a countable indexed family $\left \{ { E_j } \right \}_{j \mathop \in \N}$ of arbitrary sets in $\sigma\left({\mathcal E}\right)$.

Using the Axiom of Choice, we can create a choice function:


 * $c: \N \to \Omega$:


 * $c(j) = \alpha_j \text{ if } E_j \in \mathcal E_{\alpha_j}$

THE ABOVE NEEDS WORK: choice might not be enough, because a priori maybe there is no set $\mathcal E_{\alpha_j} \ni E_j$.

Then $\left\{ {\alpha_1, \alpha_2, \ldots} \right\}$ is countable, because $\left \{ { E_j } \right \}$ is indexed by $\N$.

By Countable Subset of Set of Countable Ordinals Has Upper Bound, this set has an upper bound. Call it $\mathcal A$.

Hence $E_j \in \mathcal E_{\mathcal A}$ for all $j \in \N$ under this construction.

Let $\mathcal B$ succeed $\mathcal A$.

Then $\displaystyle \bigcup_{j \mathop \in \N} \in \mathcal E_{\mathcal B}$.

As the sets in $\left \{ { E_j } \right \}$ were chosen arbitrarily, $\sigma\left({\mathcal E}\right) \subseteq \mathcal E_\Omega$.

AoC

Theorem
Let $\mathcal E$ such that:


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

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

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