Characterization of Euclidean Borel Sigma-Algebra

Theorem
Let $\mathcal{O}^n$, $\mathcal{C}^n$ and $\mathcal{K}^n$ be the open, closed and compact subsets of the Euclidean space $\left({\R^n, \tau}\right)$, respectively.

Let $\mathcal{J}_{ho}^n$ be the collection of half-open rectangles in $\R^n$.

Let $\mathcal{J}^n_{ho, \text{rat}} = \mathcal{J}_{ho}^n \cap \Q^n$ be the collection of half-open rectangles in $\R^n$ with rational endpoints.

Then the Borel $\sigma$-algebra $\mathcal B \left({\R^n}\right)$ satisfies:


 * $\mathcal B \left({\R^n}\right) = \sigma \left({\mathcal{O}^n}\right) = \sigma \left({\mathcal{C}^n}\right) = \sigma \left({\mathcal{K}^n}\right) = \sigma \left({\mathcal{J}_{ho}^n}\right) = \sigma \left({\mathcal{J}^n_{ho, \text{rat}}}\right)$

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

Proof
By definition of Borel $\sigma$-algebra, $\mathcal B \left({\R^n}\right) = \sigma \left({\mathcal{O}^n}\right)$.

The rest of the proof will be split in proving the following equalities:


 * $(1): \quad \sigma \left({\mathcal{O}^n}\right) = \sigma \left({\mathcal{C}^n}\right)$
 * $(2): \quad \sigma \left({\mathcal{C}^n}\right) = \sigma \left({\mathcal{K}^n}\right)$
 * $(3): \quad \sigma \left({\mathcal{O}^n}\right) = \sigma \left({\mathcal{J}_{ho}^n}\right)$
 * $(4): \quad \sigma \left({\mathcal{J}_{ho}^n}\right) = \sigma \left({\mathcal{J}^n_{ho, \text{rat}}}\right)$

Proof of $(1)$
Recall that a closed set is by definition the relative complement of an open set.

Hence Sigma-Algebra Generated by Complements of Generators applies to yield $(1)$ immediately.

Proof of $(2)$
By the Heine-Borel Theorem (Special Case), $\mathcal{K}^n \subseteq \mathcal{C}^n$.

Thus from Generated Sigma-Algebra Preserves Subset, $\sigma \left({\mathcal{K}^n}\right) \subseteq \sigma \left({\mathcal{C}^n}\right)$.

Next let, for all $n \in \N$, $B^- \left({\mathbf 0; n}\right)$ be the closed ball of radius $n$ around $\mathbf 0$ in $\R^n$.

Observe that $\R^n = \displaystyle \bigcup_{n \mathop \in \N} B^- \left({\mathbf 0; n}\right)$.

Now let $U \in \mathcal{C}^n$ be a closed subset of $\R^n$.

Then from Intersection with Subset is Subset and Intersection Distributes over Union:


 * $\displaystyle U = U \cap \R^n = U \cap \bigcup_{n \mathop \in \N} B^- \left({\mathbf 0; n}\right) = \bigcup_{n \mathop \in \N} \left({U \cap B^- \left({\mathbf 0; n}\right)}\right)$

From Intersection of Closed Sets is Closed, $U \cap B^- \left({\mathbf 0; n}\right)$ is closed for all $n \in \N$.

By definition, $B^- \left({\mathbf 0; n}\right)$ is bounded.

Thus, by the Heine-Borel Theorem, $U \cap B^- \left({\mathbf 0; n}\right)$ is compact.

Thus, any closed set is the countable union of compact sets.

By the third axiom for a $\sigma$-algebra, this means that $\mathcal{C}^n \subseteq \sigma \left({\mathcal{K}^n}\right)$.

Now the definition of generated $\sigma$-algebra ensures that $\sigma \left({\mathcal{C}^n}\right) \subseteq \sigma \left({\mathcal{K}^n}\right)$.

Hence statement $(2)$, by definition of set equality.

Proof of $(4)$
From Generated Sigma-Algebra Preserves Subset, $\sigma \left({\mathcal{J}_{ho, \text{rat}}^n}\right) \subseteq \sigma \left({\mathcal{J}_{ho}^n}\right)$.

For the converse, it will suffice to show:


 * $\mathcal{J}_{ho}^n \subseteq \sigma \left({\mathcal{J}_{ho, \text{rat}}^n}\right)$

by definition of generated $\sigma$-algebra.

So let $\lefthalf-open $n$-rectangle.

Let $\left({\mathbf{a}_m}\right)_{m \in \N}$ be a sequence in $\Q^n$ with limit $\mathbf a$.

Also, let this sequence be such that $m_1 > m_2 \implies \mathbf{a}_{m_1} > \mathbf{a}_{m_2}$, in the component-wise ordering.

Also, choose $\mathbf{b}' \in \Q^n$ such that $\mathbf{b}' > \mathbf b$, again in the component-wise ordering.

Then, for any $m \in \N$, $\left[\left[{\mathbf{a}_m \,.\,.\, \mathbf{b}'}\right)\right) \in \mathcal{J}_{ho, \text{rat}}^n$.

By Sigma-Algebra Closed Under Countable Intersection, it follows that:


 * $\displaystyle \bigcap_{m \mathop \in \N} \left[\left[{\mathbf{a}_m \,.\,.\, \mathbf{b}'}\right)\right) \in \sigma \left({\mathcal{J}_{ho, \text{rat}}^n}\right)$

Now observe, for $\mathbf x \in \R^n$:

Next, let $\left({\mathbf{b}_m}\right)_{m \in \N}$ be an increasing sequence in $\Q^n$ with limit $\mathbf{b}$.

Also, let $\mathbf{a}' \in \Q^n$ be such that $\mathbf{a}' < \mathbf a$.

Again, it follows that $\left[\left[{\mathbf{a}' \,.\,.\, \mathbf{b}_m}\right)\right) \in \mathcal{J}_{ho, \text{rat}}^n$.

A similar reasoning shows that:


 * $\displaystyle \bigcup_{m \mathop \in \N} \left[\left[{\mathbf{a}' \,.\,.\, \mathbf{b}_m}\right)\right) = \left[\left[{\mathbf{a}' \,.\,.\, \mathbf b}\right)\right)$

and that it is in $\sigma \left({\mathcal{J}^n_{ho, \text{rat}}}\right)$.

Hence by Sigma-Algebra Closed under Intersection:


 * $\left[\left[{\mathbf a \,.\,.\, \mathbf{b}'}\right)\right) \cap \left[\left[{\mathbf{a}' \,.\,.\, \mathbf b}\right)\right) \in \sigma \left({\mathcal{J}^n_{ho, \text{rat}}}\right)$

and finally (the proof of) Half-Open Rectangles Closed under Intersection yields:


 * $\left[\left[{\mathbf a \,.\,.\, \mathbf{b}'}\right)\right) \cap \left[\left[{\mathbf{a}' \,.\,.\, \mathbf b}\right)\right) = \left[\left[{\mathbf a \,.\,.\, \mathbf b}\right)\right)$

since $\mathbf{a}' < \mathbf a$ and $\mathbf b < \mathbf{b}'$, thus finishing the proof.