Inductive Construction of Sigma-Algebra Generated by Collection of Subsets

Theorem
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:


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

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


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

From $\sigma$-Algebra of Countable Sets, if $\beta$ immediately precedes $\alpha$, 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 Set Union Preserves Subsets:General Result:


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

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

Step 2
We identify numbers in $\N$ with the finite ordinals, using the definition of ordinals as initial segments.

This identification is justified from Minimal Infinite Successor Set forms Peano Structure.

Thus every $j \in \N$ can be treated as a finite initial segment of $\Omega$.

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.

First note that:


 * $\mathcal E = \mathcal E_1 \in \mathcal E_{\Omega}$

so $\mathcal E_{\Omega}$ has a unit.

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:


 * $\left\{ {\alpha_i} \right\}_{i \mathop \in \N} = \left \{ { \alpha_1, \alpha_2, \alpha_3, \ldots } \right\}$

is countable, because it is indexed by $\N$.

By Countable Subset of Minimal Uncountable Well-Ordered Set Has Upper Bound, $\left\{ {\alpha_i} \right\}$ has an upper bound. Call this bound $\gamma$.

Then $\mathcal E_{\gamma}$ is an initial segment.

Also, $E_j \in \mathcal E_{\gamma}$ for all $j \in \N$, by the definition of upper bound.

For any $\delta$ strictly succeeding $\gamma$:


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

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


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

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

We have that:


 * $\mathcal E_{\Omega}$ has a unit


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


 * $\mathcal E_{\Omega}$ is closed under countable union.

By the definition of $\sigma$-algebra, $\mathcal E_{\Omega}$ is indeed a $\sigma$-algebra.

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


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

The result follows from Part 1 and Part 2 of the proof combined, by the definition of set equality.