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

Corollary

Let the cardinality of $\EE$ satisfy:

$\size \N \le \size \EE \le \mathfrak c$

where $\mathfrak c$ denotes the cardinality of the continuum.

Then:

$\size {\map \sigma \EE} = \mathfrak c$

Proof

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

$\Box$

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.

$\blacksquare$

Axiom of Choice

This theorem depends on the Axiom of Choice, by way of Countable Union of Countable Sets is Countable.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.

Sources

• 1984: Gerald B. Folland: Real Analysis: Modern Techniques and their Applications : $\S 1.23$