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

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 $\mathbf x \in \R^n$:
{{begin-eqn}}
{{eqn | l = \mathbf x
      | o = \in 
      | r = \bigcap_{m \mathop \in \N} \horectr {\mathbf a_m} {\mathbf b'}
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall m \in \N
      | l = \mathbf x
      | o = \in
      | r = \horectr {\mathbf a_m} {\mathbf b'}
      | c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall m \in \N
      | l = \mathbf x
      | o = \ge
      | r = \mathbf a_m
      | c = {{Defof|Half-Open Rectangle}}
}}
{{eqn | lo= \land
      | l = \mathbf x
      | o = <
      | r = \mathbf b'
}}
{{eqn | ll= \leadstoandfrom
      | l = \mathbf x
      | o = \ge
      | r = \mathbf a
      | c = $\sequence {\mathbf a_m}_{m \mathop \in \N}$ is increasing with limit $\mathbf a$
}}
{{eqn | lo= \land
      | l = \mathbf x
      | o = <
      | r = \mathbf b'
}}
{{eqn | ll= \leadstoandfrom
      | l = \mathbf x
      | o = \in
      | r = \horectr {\mathbf a} {\mathbf b'}
      | c = {{Defof|Half-Open Rectangle}}
}}
{{end-eqn}}
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 $\mathbf x \in \R^n$:
{{begin-eqn}}
{{eqn | l = \mathbf x
      | o = \in
      | r = \bigcup_{m \mathop \in \N} \horectr {\mathbf a'} {\mathbf b_m}
}}
{{eqn | ll= \leadstoandfrom
      | q = \exists m \in \N
      | l = \mathbf x
      | o = \in 
      | r = \horectr {\mathbf a'} {\mathbf b_m}
      | c = {{Defof|Set Union}}
}}
{{eqn | ll= \leadstoandfrom
      | q = \exists m \in \N
      | l = \mathbf x
      | o = <
      | r = \mathbf b_m
      | c = {{Defof|Half-Open Rectangle}}
}}
{{eqn | lo= \land
      | l = \mathbf x
      | o = \ge
      | r = \mathbf a'
}}
{{eqn | ll= \leadstoandfrom
      | l = \mathbf x
      | o = <
      | r = \mathbf b
      | c = $\sequence {\mathbf b_m}_{m \mathop \in \N}$ is increasing with limit $\mathbf b$
}}
{{eqn | lo= \land
      | l = \mathbf x
      | o = \ge
      | r = \mathbf a'
}}
{{eqn | ll= \leadstoandfrom
      | l = \mathbf x
      | o = \in
      | r = \horectr {\mathbf a'} {\mathbf b}
      | c = {{Defof|Half-Open Rectangle}}
}}
{{end-eqn}}
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.
{{qed}}