A box in ${\mathbb{R}}^n$ with parameters $a_i\le b_i$ is Lebesgue measurable of measure ${\prod }_{i<n}(b_i-a_i)$, whichever of its faces are included

$R^{\circ }$ is open and $\bar{R}$ is closed in $({\mathbb{R}}^n,d_2)$, by the same coordinatewise estimate in each case, so both are Borel and hence Lebesgue measurable.

A closed rectangle with a degenerate side is Lebesgue null: let $u_i\le v_i$ be reals with $u_{i_0}=v_{i_0}=c$ for some $i_0<n$, and let $\eta$ be a positive real; the half-open box with parameter pairs $(u_i-1,v_i]$ for $i\mathrm{\ne }{i}_0$ and $(c-\eta ,c]$ in coordinate $i_0$ is nonempty, contains $[u,v]$, and has volume $\eta C$ where $C:={\prod }_{i<n}{w}_i>0$ with $w_{i_0}:=1$ and $w_i:=v_i-u_i+1$ otherwise, so monotonicity of the outer measure gives ${\lambda }_n^{\ast }([u,v])\le \eta C$ for every positive real $\eta$ and hence ${\lambda }_n^{\ast }([u,v])=0$.

The difference $\bar{R}\setminus R^{\circ }$ is contained in the union of the $2n$ closed rectangles obtained from $[a,b]$ by replacing the $i$-th side by the degenerate side $[a_i,a_i]$ or by $[b_i,b_i]$, each of which is Lebesgue null by step 1.2, so countable subadditivity of the outer measure, applied to that finite list padded with empty sets, gives ${\lambda }_n^{\ast }(\bar{R}\setminus R^{\circ })=0$; every subset of $\bar{R}\setminus R^{\circ }$ is therefore Lebesgue measurable of measure $0$.

Suppose instead $a_{i_0}=b_{i_0}$ for some $i_0<n$. Then $R^{\circ }=\mathrm{\varnothing }$, the rectangle $\bar{R}$ is Lebesgue null by step 1.2, every $R$ between them is a subset of it and so is measurable of measure $0$, and the product ${\prod }_{i<n}(b_i-a_i)$ has the factor $0$ and is therefore $0$ as well.

Suppose first that $a_i<b_i$ for every $i<n$. Then $B(a,b)$ is a nonempty half-open box with $R^{\circ }\subseteq B(a,b)\subseteq \bar{R}$ and ${\lambda }_n(B(a,b))=\mathrm{vol}(B(a,b))={\prod }_{i<n}(b_i-a_i)$. For $R$ with $R^{\circ }\subseteq R\subseteq \bar{R}$, both $R\setminus B(a,b)$ and $B(a,b)\setminus R$ are contained in $\bar{R}\setminus R^{\circ }$, hence are measurable of measure $0$ by step 2.1, so $R=(B(a,b)\setminus (B(a,b)\setminus R))\cup (R\setminus B(a,b))$ is measurable, and additivity on the two disjoint decompositions $R=(R\cap B(a,b))\bigsqcup (R\setminus B(a,b))$ and $B(a,b)=(R\cap B(a,b))\bigsqcup (B(a,b)\setminus R)$ gives ${\lambda }_n(R)={\lambda }_n(R\cap B(a,b))={\lambda }_n(B(a,b))$.

Steps 3.1 and 2.2 exhaust the two cases and give the displayed value in each, and step 1.1 supplies the Borel and measurability clauses for $R^{\circ }$ and $\bar{R}$.