Inductive Construction of Sigma-Algebra Generated by Collection of Subsets

Theorem
Let $\EE$ be a set of sets which are subsets of some set $X$.

Let $\map \sigma \EE$ be the $\sigma$-algebra generated by $\EE$.

Then $\map \sigma \EE$ 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:


 * $\EE_1 = \EE$


 * $\EE_\alpha = \begin{cases} \set {\SS \in \powerset {\EE_\beta}: \SS \text { is countable or } \SS^\complement \text{ is countable} } & \alpha \text{ has an immediate predecessor } \beta \\

\ds \bigcup_{\beta \mathop \prec \alpha} \EE_\beta & \text { otherwise} \end{cases}$


 * $\EE_\Omega = \ds \bigcup_{\alpha \mathop \in \Omega} \EE_\alpha$

Then $\map \sigma \EE = \EE_\Omega$.

Step 1
We will show that $\map \sigma \EE \subseteq \EE_\Omega$.

Define:


 * $\OO = \set {o \in \Omega: \EE_o \in \map \sigma \EE}$

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


 * $\EE_1 \subseteq \map \sigma \EE$.

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

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

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

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

Thus $\EE_\alpha \subseteq \map \sigma \EE$ for all $\alpha \in \Omega$.

By Set Union Preserves Subsets:General Result:


 * $\ds \bigcup_{\alpha \mathop \in \Omega} \EE_\alpha \subseteq \map \sigma \EE$

Thus $\EE_{\Omega} \subseteq \map \sigma \EE$.

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

This identification is justified from Minimally Inductive 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 $\EE$, the reverse inclusion:


 * $\EE_\Omega \supseteq \map \sigma \EE$

will follow if $\EE_\Omega$ is a $\sigma$-algebra.

First note that:


 * $\EE = \EE_1 \in \EE_\Omega$

so $\EE_\Omega$ has a unit.

Let:


 * $\family {E_j}_{j \mathop \in \N}$

be an arbitrary countable indexed family of sets in $\EE_\Omega$.

By the inductive construction of $\EE_\Omega$, for all $j \in \N$:


 * $E_j, E_j^\complement \in \EE_{\alpha_j}$

for some initial segment $\alpha_j$,

Thus $\EE_\Omega$ is closed under complement.

The set:


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

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

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

Call this bound $\gamma$.

Then $\EE_\gamma$ is an initial segment.

Also, $E_j \in \EE_\gamma$ for all $j \in \N$, by the definition of upper bound.

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


 * $\ds \bigcup_{j \mathop \in \N} E_j \in \EE_\delta$

Thus by the construction of $\EE_\Omega$:


 * $\EE_\delta \subseteq \EE_\Omega$

This means that $\EE_\Omega$ is closed under countable union.

We have that:


 * $\EE_\Omega$ has a unit


 * $\EE_\Omega$ is closed under complement.


 * $\EE_\Omega$ is closed under countable union.

By the definition of $\sigma$-algebra, $\EE_\Omega$ is indeed a $\sigma$-algebra.

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


 * $\map \sigma \EE \subseteq \EE_\Omega$

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