Generators for Extended Real Sigma-Algebra

Theorem
Let $\overline{\mathcal B}$ be the extended real $\sigma$-algebra.

Then $\overline{\mathcal B}$ is generated by each of the following collections of extended real intervals:

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

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

Furthermore, write $\Sigma_i$ for $\sigma \left({\mathcal G_i}\right)$ and $\Sigma'_i$ for $\sigma \left({\mathcal G'_i}\right)$.

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 $a \in \R$):


 * $\complement_{\overline \R} \left({\left[{a \,.\,.\, +\infty}\right]}\right) = \left[{-\infty \,.\,.\, a}\right)$
 * $\complement_{\overline \R} \left({\left[{-\infty \,.\,.\, a}\right]}\right) = \left({a \,.\,.\, +\infty}\right]$

we deduce that:


 * $\mathcal G_3 \subseteq \Sigma_1, \mathcal G'_3 \subseteq \Sigma'_1$
 * $\mathcal G_2 \subseteq \Sigma_4, \mathcal G'_2 \subseteq \Sigma'_4$

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 $a \in \R$:


 * $\left[{a \,.\,.\, +\infty}\right] = \displaystyle \bigcap_{n \mathop \in \N} \left({a - \frac 1 n, +\infty}\right]$

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


 * $\mathcal G_1 \subseteq \Sigma_2, \mathcal G'_1, \subseteq \Sigma'_2$

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


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

For the converse inclusion, remark that:


 * $\left({a \,.\,.\, +\infty}\right] = \displaystyle \bigcup_{n \mathop \in \N} \left[{a + \frac 1 n, +\infty}\right]$

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:


 * $\left({a \,.\,.\, +\infty}\right] = \displaystyle \bigcup_{\substack{q \mathop \in \Q \\ q \mathop > a}} \left({q \,.\,.\, +\infty}\right]$

whence $\Sigma_2 \subseteq \Sigma'_2$, and all eight $\sigma$-algebras are equal.

It remains to establish that in fact they equal $\overline{\mathcal B}$.

Since all elements of $\mathcal G_2$ are open in the extended real number space, it follows that:


 * $\Sigma_2 \subseteq \overline{\mathcal B}$