The Euclidean Lebesgue measure is the completion of the product of the factor Lebesgue measures

Let $A\in \mathcal{L}({\mathbb{R}}^m)$ and $B\in \mathcal{L}({\mathbb{R}}^n)$. By [L2], choose Borel cores $A_0,B_0$ and Borel null hulls $Z,W$ such that $A\mathrm{△}{A}_0\subseteq Z$ and $B\mathrm{△}{B}_0\subseteq W$. The slabs $Z\times {\mathbb{R}}^n$ and ${\mathbb{R}}^m\times W$ are Euclidean null: for example, $Z\times {\mathbb{R}}^n={\bigcup }_{k\ge 1}Z\times [-k,k{]}^n$, and [L1], [L3], and [L4] give Euclidean measure $0$ to each Borel rectangle in this union. Since $(A\times B)\mathrm{△}(A_0\times B_0)\subseteq (Z\times {\mathbb{R}}^n)\cup ({\mathbb{R}}^m\times W),$ [L2] makes $A\times B$ Euclidean Lebesgue measurable.

Step 1.1 puts every measurable rectangle in $\mathcal{L}({\mathbb{R}}^{m+n})$, so $\mathcal{L}({\mathbb{R}}^m)\otimes \mathcal{L}({\mathbb{R}}^n)\subseteq \mathcal{L}({\mathbb{R}}^{m+n})$. Let $\rho$ be the restriction of ${\lambda }_{m+n}$ to this product sigma-algebra. For the rectangle in step 1.1, completeness [L5] and the null symmetric difference give $\rho (A\times B)={\lambda }_{m+n}(A_0\times B_0)$; [L1] and [L2] identify this with ${\lambda }_m(A){\lambda }_n(B)$. Thus $\rho$ has the product rectangle formula. By [L4] the factors are sigma-finite, so uniqueness in [L3] gives $\rho ={\lambda }_m\times {\lambda }_n$.

Because the product sigma-algebra is contained in the complete Euclidean Lebesgue sigma-algebra and the measures agree there by step 2.1, its completion is contained in $\mathcal{L}({\mathbb{R}}^{m+n})$. Conversely, if $E\in \mathcal{L}({\mathbb{R}}^{m+n})$, [L2] gives a Borel $C$ and a Borel null set $Z$ with $E\mathrm{△}C\subseteq Z$. The Borel sets $C,Z$ belong to the product sigma-algebra, and [L1] and step 2.1 give $({\lambda }_m\times {\lambda }_n)(Z)={\lambda }_{m+n}(Z)=0$. Hence $E$ belongs to the completion of the product measure. The domains and measures therefore coincide, which is exactly the completion claim of [L6].