Characterization of Euclidean Borel Sigma-Algebra

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Theorem

Let $\mathcal{O}^n$, $\mathcal{C}^n$ and $\mathcal{K}^n$ be the collections of 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:

$\sigma \left({\mathcal O^n}\right) = \sigma \left({\mathcal C^n}\right)$

$\blacksquare$


Proof of $(2)$

By the Heine-Borel Theorem, $\mathcal K^n \subseteq \mathcal C^n$.

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 the result, by definition of set equality.

$\blacksquare$


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 Characterization of Euclidean Borel Sigma-Algebra/Open equals Closed, 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 is Countable.


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:

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

$\blacksquare$


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 $\left[[{\mathbf a \,.\,.\, \mathbf b}\right))$ be a half-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 $\mathbf x \in \R^n$:

\(\displaystyle \mathbf x\) \(\in\) \(\displaystyle \bigcap_{m \mathop \in \N} \left[\left[{\mathbf a_m \,.\,.\, \mathbf b'}\right)\right)\)
\(\displaystyle \iff \ \ \) \(\displaystyle \forall m \in \N: \mathbf x\) \(\in\) \(\displaystyle \left[\left[{\mathbf a_m \,.\,.\, \mathbf b'}\right)\right)\) Definition of Set Intersection
\(\displaystyle \iff \ \ \) \(\displaystyle \forall m \in \N: \mathbf x\) \(\ge\) \(\displaystyle \mathbf a_m \land \mathbf x < \mathbf b'\) Definition of Half-Open Rectangle
\(\displaystyle \iff \ \ \) \(\displaystyle \mathbf x\) \(\ge\) \(\displaystyle \mathbf a \land \mathbf x < \mathbf b'\) $\left({\mathbf a_m}\right)_{m \in \N}$ is increasing with limit $\mathbf a$
\(\displaystyle \iff \ \ \) \(\displaystyle \mathbf x\) \(\in\) \(\displaystyle \left[\left[{\mathbf a \,.\,.\, \mathbf b'}\right)\right)\) Definition of Half-Open Rectangle


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 $\mathbf x \in \R^n$:

\(\displaystyle \mathbf x\) \(\in\) \(\displaystyle \bigcup_{m \mathop \in \N} \left[\left[{\mathbf a' \,.\,.\, \mathbf b_m}\right)\right)\)
\(\displaystyle \iff \ \ \) \(\displaystyle \exists m \in \N: \mathbf x\) \(\in\) \(\displaystyle \left[\left[{\mathbf a' \,.\,.\, \mathbf b_m}\right)\right)\) Definition of Set Union
\(\displaystyle \iff \ \ \) \(\displaystyle \exists m \in \N: \mathbf x\) \(<\) \(\displaystyle \mathbf b_m \land \mathbf x \ge \mathbf a'\) Definition of Half-Open Rectangle
\(\displaystyle \iff \ \ \) \(\displaystyle \mathbf x\) \(<\) \(\displaystyle \mathbf b \land \mathbf x \ge \mathbf a'\) $\left({\mathbf b_m}\right)_{m \in \N}$ is increasing with limit $\mathbf b$
\(\displaystyle \iff \ \ \) \(\displaystyle \mathbf x\) \(\in\) \(\displaystyle \left[\left[{\mathbf a' \,.\,.\, \mathbf b}\right)\right)\) Definition of Half-Open Rectangle

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.

$\blacksquare$


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