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