Characterization of Euclidean Borel Sigma-Algebra

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Theorem

Let $\OO^n$, $\CC^n$ and $\KK^n$ be the collections of open, closed and compact subsets of the Euclidean space $\struct {\R^n, \tau}$, respectively.

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

Let $\JJ^n_{ho, \text {rat} }$ be the collection of half-open rectangles in $\R^n$ with rational endpoints.


Then the Borel $\sigma$-algebra $\map \BB {\R^n}$ satisfies:

$\map \BB {\R^n} = \map \sigma {\OO^n} = \map \sigma {\CC^n} = \map \sigma {\KK^n} = \map \sigma {\JJ_{ho}^n} = \map \sigma {\JJ^n_{ho, \text {rat} } }$

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


Proof

By definition of Borel $\sigma$-algebra, $\map \BB {\R^n} = \map \sigma {\OO^n}$.

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

$(1): \quad \map \sigma {\OO^n} = \map \sigma {\CC^n}$
$(2): \quad \map \sigma {\CC^n} = \map \sigma {\KK^n}$
$(3): \quad \map \sigma {\OO^n} = \map \sigma {\JJ_{ho}^n}$
$(4): \quad \map \sigma {\JJ_{ho}^n} = \map \sigma {\JJ^n_{ho, \text {rat} } }$


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:

$\map \sigma {\OO^n} = \map \sigma {\CC^n}$

$\blacksquare$


Proof of $(2)$

By the Heine-Borel Theorem, $\KK^n \subseteq \CC^n$.

Thus from Generated Sigma-Algebra Preserves Subset, $\map \sigma {\KK^n} \subseteq \map \sigma {\CC^n}$.


Next, let, for all $n \in \N$, $\map {B_n^-} {\mathbf 0}$ be the closed ball of radius $n$ around $\mathbf 0$ in $\R^n$.

Observe that $\R^n = \ds \bigcup_{n \mathop \in \N} \map {B_n^-} {\mathbf 0}$.


Now let $U \in \CC^n$ be a closed subset of $\R^n$.

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

$\ds U = U \cap \R^n = U \cap \bigcup_{n \mathop \in \N} \map {B_n^-} {\mathbf 0} = \bigcup_{n \mathop \in \N} \paren {U \cap \map {B_n^-} {\mathbf 0} }$

From Intersection of Closed Sets is Closed in Topological Space, $U \cap \map {B_n^-} {\mathbf 0}$ is closed for all $n \in \N$.

By definition, $\map {B_n^-} {\mathbf 0}$ is bounded.

Thus, by the Heine-Borel Theorem, $U \cap \map {B_n^-} {\mathbf 0}$ is compact.


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

By the third axiom for a $\sigma$-algebra, this means that $\CC^n \subseteq \map \sigma {\KK^n}$.

Now the definition of generated $\sigma$-algebra ensures that $\map \sigma {\CC^n} \subseteq \map \sigma {\KK^n}$.


Hence the result, by definition of set equality.

$\blacksquare$


Proof of $(3)$

Let $\horectr {\mathbf a} {\mathbf b} \in \JJ^n_{ho}$.

Then:

$\horectr {\mathbf a} {\mathbf b} = \paren {\openint \gets {\mathbf b} } \cap \horectr {\mathbf a} \to$

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 $\map \BB {\R^n}$, and so Sigma-Algebra Closed under Intersection yields:

$\horectr {\mathbf a} {\mathbf b} \in \map \sigma {\OO^n}$

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

$\map \sigma {\JJ^n_{ho} } \subseteq \map \sigma {\OO^n}$


Denote $\mathbf 1 = \tuple {1, \ldots, 1} \in \R^n$.

Define, for all $k \in \N$, $\map \SS k$ by:

$\map \SS k := \set {\horectr {2^{-k} \mathbf j} {2^{-k} \paren {\mathbf j + \mathbf 1} } : \mathbf j \in \Z^n}$

It is immediate that $\ds \bigcup \map \SS k = \R^n$ and $\map \SS k \subseteq \JJ^n_{ho}$.

Also, $\map \SS k$ is countable from Cartesian Product of Countable Sets is Countable.


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

$\ds U_k := \bigcup \set {S \in \map \SS k: S \subseteq U}$

From Set Union Preserves Subsets:

$U_k \subseteq U$


Also, $U_k \in \map \sigma {\JJ^n_{ho} }$ since the union is countable.

It follows that also $\ds \bigcup_{k \mathop \in \N} U_k \in \map \sigma {\JJ^n_{ho} }$.

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

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

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

As $U$ is open, there exists an $\epsilon > 0$ such that the open ball $\map {B_\epsilon} {\mathbf x}$ is contained in $U$.

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

$\mathbf x \in \horectr {2^{-k} \mathbf j} {2^{-k} \paren {\mathbf j + \mathbf 1} }$

Now it is to be shown that:

$\horectr {2^{-k} \mathbf j} {2^{-k} \paren {\mathbf j + \mathbf 1} } \subseteq \map {B_\epsilon} {\mathbf x}$

To this end, observe that for any $\mathbf y \in \horectr {2^{-k} \mathbf j} {2^{-k} \paren {\mathbf j + \mathbf 1} }$, it holds that:

$\map d {\mathbf x, \mathbf y} \le \map \diam {\horectr {2^{-k} \mathbf j} {2^{-k} \paren {\mathbf j + \mathbf 1} } }$

by definition of diameter.

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

$\norm {2^{-k} \paren {\mathbf j + \mathbf 1} - 2^{-k} \mathbf j} = \norm {2^{-k} \mathbf 1} = \sqrt n \, 2^{-k}$

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


Hence:

$\horectr {2^{-k} \mathbf j} {2^{-k} \paren {\mathbf j + \mathbf 1} } \subseteq \map {B_\epsilon} {\mathbf x}$

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

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


Thereby we have shown that:

$\map \sigma {\JJ^n_{ho} } = \map \sigma {\OO^n}$

$\blacksquare$


Proof of $(4)$

From Generated Sigma-Algebra Preserves Subset:

$\map \sigma {\JJ_{ho, \text{rat} }^n} \subseteq \map \sigma {\JJ_{ho}^n}$


For the converse, it will suffice to show:

$\JJ_{ho}^n \subseteq \map \sigma {\JJ_{ho, \text{rat} }^n}$

by definition of generated $\sigma$-algebra.


So, let $\horectr {\mathbf a} {\mathbf b}$ be a half-open $n$-rectangle.


Let $\sequence {\mathbf a_m}_{m \mathop \in \N}$ 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 $m \in \N$:

$\horectr {\mathbf a_m} {\mathbf b'} \in \JJ_{ho, \text{rat} }^n$

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

$\ds \bigcap_{m \mathop \in \N} \horectr {\mathbf a_m} {\mathbf b'} \in \map \sigma {\JJ_{ho, \text{rat} }^n}$


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

\(\ds \mathbf x\) \(\in\) \(\ds \bigcap_{m \mathop \in \N} \horectr {\mathbf a_m} {\mathbf b'}\)
\(\ds \leadstoandfrom \ \ \) \(\ds \forall m \in \N: \, \) \(\ds \mathbf x\) \(\in\) \(\ds \horectr {\mathbf a_m} {\mathbf b'}\) Definition of Set Intersection
\(\ds \leadstoandfrom \ \ \) \(\ds \forall m \in \N: \, \) \(\ds \mathbf x\) \(\ge\) \(\ds \mathbf a_m\) Definition of Half-Open Rectangle
\(\, \ds \land \, \) \(\ds \mathbf x\) \(<\) \(\ds \mathbf b'\)
\(\ds \leadstoandfrom \ \ \) \(\ds \mathbf x\) \(\ge\) \(\ds \mathbf a\) $\sequence {\mathbf a_m}_{m \mathop \in \N}$ is increasing with limit $\mathbf a$
\(\, \ds \land \, \) \(\ds \mathbf x\) \(<\) \(\ds \mathbf b'\)
\(\ds \leadstoandfrom \ \ \) \(\ds \mathbf x\) \(\in\) \(\ds \horectr {\mathbf a} {\mathbf b'}\) Definition of Half-Open Rectangle


Next, let $\sequence {\mathbf b_m}_{m \mathop \in \N}$ 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 $\horectr {\mathbf a'} {\mathbf b_m} \in \JJ_{ho, \text{rat} }^n$.

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

$\ds \bigcup_{m \mathop \in \N} \horectr {\mathbf a'} {\mathbf b_m} \in \map \sigma {\JJ^n_{ho} }$


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

\(\ds \mathbf x\) \(\in\) \(\ds \bigcup_{m \mathop \in \N} \horectr {\mathbf a'} {\mathbf b_m}\)
\(\ds \leadstoandfrom \ \ \) \(\ds \exists m \in \N: \, \) \(\ds \mathbf x\) \(\in\) \(\ds \horectr {\mathbf a'} {\mathbf b_m}\) Definition of Set Union
\(\ds \leadstoandfrom \ \ \) \(\ds \exists m \in \N: \, \) \(\ds \mathbf x\) \(<\) \(\ds \mathbf b_m\) Definition of Half-Open Rectangle
\(\, \ds \land \, \) \(\ds \mathbf x\) \(\ge\) \(\ds \mathbf a'\)
\(\ds \leadstoandfrom \ \ \) \(\ds \mathbf x\) \(<\) \(\ds \mathbf b\) $\sequence {\mathbf b_m}_{m \mathop \in \N}$ is increasing with limit $\mathbf b$
\(\, \ds \land \, \) \(\ds \mathbf x\) \(\ge\) \(\ds \mathbf a'\)
\(\ds \leadstoandfrom \ \ \) \(\ds \mathbf x\) \(\in\) \(\ds \horectr {\mathbf a'} {\mathbf b}\) Definition of Half-Open Rectangle

Hence, it follows that:

$\ds \bigcup_{m \mathop \in \N} \horectr {\mathbf a'} {\mathbf b_m} = \horectr {\mathbf a'} {\mathbf b}$

whence the latter is in $\map \sigma {\JJ^n_{ho, \text{rat} } }$.


Hence by Sigma-Algebra Closed under Intersection:

$\horectr {\mathbf a} {\mathbf b'} \cap \horectr {\mathbf a'} {\mathbf b} \in \map \sigma {\JJ^n_{ho, \text{rat} } }$

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

$\horectr {\mathbf a} {\mathbf b'} \cap \horectr {\mathbf a'} {\mathbf b} = \horectr {\mathbf a} {\mathbf b}$

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

$\blacksquare$


Sources

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