On Borel subsets of R^{m+n}, the product lambda_m times lambda_n agrees with lambda_{m+n}

Given: The Axiom of Countable Choice, positive integers $m,n$, and the identification ${\mathbb{R}}^{m+n}={\mathbb{R}}^m\times {\mathbb{R}}^n$.

The Borel sigma-algebra on ${\mathbb{R}}^{m+n}$ is $\mathcal{B}({\mathbb{R}}^m)\otimes \mathcal{B}({\mathbb{R}}^n)$. (The Borel product of R^m and R^n is the Borel sigma-algebra of R^{m+n})

The product measure on sigma-finite spaces exists and satisfies the rectangle formula. (For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique)

Assuming countable choice, Lebesgue measure of a box is the product of its side lengths. (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)

Two measures that agree on a sigma-finite generating pi-system agree on the generated sigma-algebra. (Measures agreeing on a generating pi-system are equal under an increasing finite-measure exhaustion from that pi-system)

Rational half-open boxes, together with the empty set, form a generating pi-system for $\mathcal{B}({\mathbb{R}}^{m+n})$. The increasing boxes $Q_N=(-N,N{]}^{m+n}$ exhaust the space and have finite measure for both measures in the Statement.