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

Theorem
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:


 * $\sigma \left({\mathcal{J}_{ho}^n}\right) = \sigma \left({\mathcal{J}^n_{ho, \text{rat}}}\right)$

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

Proof
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.