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 $\mathcal{J}^n_{ho, \text{rat}}$ 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 $(3)$
Let $\left[[{\mathbf a \,.\,.\, \mathbf b}\right)) \in \mathcal{J}^n_{ho}$.

Then:


 * $\left[[{\mathbf a \,.\,.\, \mathbf b}\right)) = \left(({-\infty \,.\,.\, \mathbf b}\right)) \cap \left[[{\mathbf a \,.\,.\, +\infty}\right))$

provides a way of writing this half-open $n$-rectangle as an intersection of an open and a closed set.

By $(1)$, these are both in $\mathcal B \left({\R^n}\right)$, and so Sigma-Algebra Closed under Intersection yields:


 * $\left[[{\mathbf a \,.\,.\, \mathbf b}\right)) \in \sigma \left({\mathcal{O}^n}\right)$

Hence, by definition of generated $\sigma$-algebra:


 * $\sigma \left({\mathcal{J}^n_{ho}}\right) \subseteq \sigma \left({\mathcal{O}^n}\right)$

Denote $\mathbf 1 = \left({1, \ldots, 1}\right) \in \R^n$.

Define, for all $k \in \N$, $\mathcal S \left({k}\right)$ by:


 * $\mathcal S \left({k}\right) := \left\{{ \left[\left[{2^{-k}\mathbf j \,.\,.\, 2^{-k} \left({\mathbf j + \mathbf 1}\right)}\right)\right) : \mathbf j \in \Z^n}\right\}$

It is immediate that $\displaystyle \bigcup \mathcal S \left({k}\right) = \R^n$ and $\mathcal S \left({k}\right) \subseteq \mathcal{J}^n_{ho}$.

Also, $\mathcal S \left({k}\right)$ is countable from Cartesian Product of Countable Sets.

Now define, again for all $k \in \N$, $U_k$ by:


 * $\displaystyle U_k := \bigcup \, \left\{{S \in \mathcal S \left({k}\right): S \subseteq U}\right\}$

From Set Union Preserves Subsets, $U_k \subseteq U$.

Also, $U_k \in \sigma \left({\mathcal{J}^n_{ho}}\right)$ since the union is countable.

It follows that also $\displaystyle \bigcup_{k \mathop \in \N} U_k \in \sigma \left({\mathcal{J}^n_{ho}}\right)$.

Next, it is to be shown that $\displaystyle \bigcup_{k \mathop \in \N} U_k = U$.

Note that Set Union Preserves Subsets ensures $\displaystyle \bigcup_{k \mathop \in \N} U_k \subseteq U$.

For the converse, let $\mathbf x \in U$.

As $U$ is open, there exists an $\epsilon > 0$ such that the open ball $B \left({\mathbf x; \epsilon}\right)$ is contained in $U$.

Fix $k \in \N$ such that $\sqrt n \, 2^{-k} < \epsilon$, and find $\mathbf j \in \Z^n$ such that:


 * $\mathbf x \in \left[\left[{2^{-k} \mathbf j \,.\,.\, 2^{-k} \left({\mathbf j + \mathbf 1}\right)}\right)\right)$

Now it is to be shown that:


 * $\left[\left[{2^{-k} \mathbf j \,.\,.\, 2^{-k} \left({\mathbf j + \mathbf 1}\right)}\right)\right) \subseteq B \left({\mathbf x; \epsilon}\right)$

To this end, observe that for any $\mathbf y \in \left[\left[{2^{-k} \mathbf j \,.\,.\, 2^{-k} \left({\mathbf j + \mathbf 1}\right)}\right)\right)$, it holds that:


 * $d \left({\mathbf x, \mathbf y}\right) \le \operatorname{diam} \left({\left[\left[{2^{-k} \mathbf j \,.\,.\, 2^{-k} \left({\mathbf j + \mathbf 1}\right)}\right)\right)}\right)$

by definition of diameter.

Now from Diameter of Rectangle, the right-hand side equals:


 * $\left\Vert{2^{-k} \left({\mathbf j + \mathbf 1}\right) - 2^{-k} \mathbf j}\right\Vert = \left\Vert{2^{-k} \mathbf 1}\right\Vert = \sqrt{n} \, 2^{-k}$

which is smaller than $\epsilon$ by the way $k$ was chosen.

Hence:


 * $\left[\left[{2^{-k} \mathbf j \,.\,.\, 2^{-k} \left({\mathbf j + \mathbf 1}\right)}\right)\right) \subseteq B \left({\mathbf x; \epsilon}\right)$

and so every $\mathbf x \in U$ is contained in some $U_k$.

Thus it follows that $U \subseteq \displaystyle \bigcup_{k \mathop \in \N} U_k$.

Thereby we have shown that:


 * $\sigma \left({\mathcal{J}^n_{ho}}\right) = \sigma \left({\mathcal{O}^n}\right)$

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

Thus, by the third axiom for a $\sigma$-algebra:


 * $\displaystyle \bigcup_{m \mathop \in \N} \left[\left[{\mathbf{a}' \,.\,.\, \mathbf{b}_m}\right)\right) \in \sigma \left({\mathcal{J}^n_{ho}}\right)$

Similar to the above approach, for any $\mathbf x \in \R^n$:

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

whence the latter 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.