Characterization of Euclidean Borel Sigma-Algebra/Rectangle equals Rational Rectangle

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Theorem

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:

$\map \sigma {\JJ_{ho}^n} = \map \sigma {\JJ^n_{ho, \text{rat} } }$

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


Proof

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.

$\blacksquare$

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