Generators for Extended Real Sigma-Algebra

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Let $\overline \BB$ be the extended real $\sigma$-algebra.

Then $\overline \BB$ is generated by each of the following collections of extended real intervals:

\(\text {(1)}: \quad\) \(\ds \) \(\) \(\ds \set {\ \closedint a \to: a \in \R}\)
\(\text {(1')}: \quad\) \(\ds \) \(\) \(\ds \set {\ \closedint a \to: a \in \Q}\)
\(\text {(2)}: \quad\) \(\ds \) \(\) \(\ds \set {\ \hointl b \to: b \in \R}\)
\(\text {(2')}: \quad\) \(\ds \) \(\) \(\ds \set {\ \hointl b \to: b \in \Q}\)
\(\text {(3)}: \quad\) \(\ds \) \(\) \(\ds \set {\ \hointr \gets c: c \in \R}\)
\(\text {(3')}: \quad\) \(\ds \) \(\) \(\ds \set {\ \hointr \gets c: c \in \Q}\)
\(\text {(4)}: \quad\) \(\ds \) \(\) \(\ds \set {\ \closedint \gets d: d \in \R}\)
\(\text {(4')}: \quad\) \(\ds \) \(\) \(\ds \set {\ \closedint \gets d: d \in \Q}\)


Let us first establish that $(1)$ up to $(4')$ all generate the same $\sigma$-algebra.

Denote $\GG_i$ for the collection at point $(i)$, and $\GG'_i$ for that at $(i')$, where $i = 1, 2, 3, 4$.

Furthermore, write $\Sigma_i$ for $\map \sigma {\GG_i}$ and $\Sigma'_i$ for $\map \sigma {\GG'_i}$.

Here $\sigma$ denotes generated $\sigma$-algebra.

By Generated Sigma-Algebra Preserves Subset, we have the following inclusions:

$\Sigma'_i \subseteq \Sigma_i$

for $i = 1, 2, 3, 4$.

Since we have the following observations about complements in $\overline \R$ (for arbitrary $a \in \R$):

$\relcomp {\overline \R} {\closedint a \to} = \hointr \gets a$
$\relcomp {\overline \R} {\closedint \gets a} = \hointl a \to$

we deduce that:

$\GG_3 \subseteq \Sigma_1, \GG'_3 \subseteq \Sigma'_1$
$\GG_2 \subseteq \Sigma_4, \GG'_2 \subseteq \Sigma'_4$

and by definition of generated $\sigma$-algebra:

$\Sigma_3 \subseteq \Sigma_1, \Sigma'_3 \subseteq \Sigma'_1$
$\Sigma_2 \subseteq \Sigma_4, \Sigma'_2 \subseteq \Sigma'_4$

By Complement of Complement, the converse inclusions:

$\Sigma_1 \subseteq \Sigma_3, \Sigma'_1 \subseteq \Sigma'_3$
$\Sigma_4 \subseteq \Sigma_2, \Sigma'_4 \subseteq \Sigma'_2$

are derived.

Subsequently, remark that, for all $a \in \R$:

$\closedint a \to = \ds \bigcap_{n \mathop \in \N_{>0}} \hointl {a - \frac 1 n} \to$

and by Sigma-Algebra Closed under Countable Intersection, it follows that:

$\GG_1 \subseteq \Sigma_2, \GG'_1, \subseteq \Sigma'_2$

whence by definition of generated $\sigma$-algebra:

$\Sigma_1 \subseteq \Sigma_2, \Sigma'_1 \subseteq \Sigma'_2$

For the converse inclusion, remark that:

$\hointl a \to = \ds \bigcup_{n \mathop \in \N_{>0}} \closedint {a + \frac 1 n} \to$

and thus immediately establish:

$\Sigma_2 \subseteq \Sigma_1, \Sigma'_2 \subseteq \Sigma'_1$

To summarize, the above arguments establish:

$\Sigma'_1 = \Sigma'_2 = \Sigma'_3 = \Sigma'_4 \subseteq \Sigma_4 = \Sigma_3 = \Sigma_2 = \Sigma_1$

Finally, for all $a \in \R$, we have:

$\hointl a \to = \ds \bigcup_{\substack {q \mathop \in \Q \\ q \mathop > a}} \hointl q \to$

whence $\Sigma_2 \subseteq \Sigma'_2$, and all eight $\sigma$-algebras are equal; denote them by $\Sigma$ from now on.

It remains to establish that in fact they equal $\overline \BB$.

Since all elements of $\GG_2$ are open sets in the extended real number space, it follows that:

$\Sigma_2 \subseteq \overline \BB$

By Intervals with Extended Rational Endpoints form Countable Basis for Extended Real Number Space, every open set in the extended real number space is of the form:

$\ds \bigcup \QQ$

where $\QQ$ is a collection of extended real intervals with extended rational endpoints.

By Set of Intervals with Extended Rational Endpoints is Countable, $\QQ$ is a subset of a countable set.

By Subset of Countably Infinite Set is Countable, $\QQ$ is countable as well.

It follows by $\sigma$-algebra axiom $(3)$ that it suffices to show that:

$\openint a b \in \Sigma$
$\hointl a \to \in \Sigma$
$\hointr \gets b \in \Sigma$
$\closedint \to \gets \in \Sigma$

for all rational numbers $a, b \in \Q$.

The middle two are in $\Sigma'_2$ and $\Sigma'_4$, respectively, hence in $\Sigma$.

By Sigma-Algebra Closed under Intersection:

$\openint a b = \hointl a \to \cap \hointr \gets b \in \Sigma$

and by Sigma-Algebra Closed under Union:

$\closedint \to \gets = \closedint a \to \cup \closedint \gets a \in \Sigma$

Hence $\Sigma = \overline \BB$, as desired.