I would like to show the Lebesgue measure $\lambda_{n + m}$ in $\mathbf{R}^{n + m}$ has the product property $$ \lambda_{n + m}(A \times B) = \lambda_n(A) \cdot \lambda_m(B) $$ for $A \in \mathcal{B}_{\mathbf{R}^n}$ and $B \in \mathcal{B}_{\mathbf{R}^m}$, where $\mathcal{B}_{\mathbf{R}^n}$ is the Borel measurable sets in $\mathbf{R}^n$. I know there are similar questions asked on the platform such as $ \lambda_{n+m}( A \times B ) = \lambda_n(A) \cdot \lambda_m(B) $. However, I am not entirely convince of the proof given in the link.
In particular, in the link the author proved the case the set $\mathcal{D}_I$ is a $\lambda$-system, with $I$ being a bounded rectangle (in fact the proof for this is not so convincing either, it seems like the author proved an identity where $B$ is contained fully in a cube $Q$ first and then used this identity in the case where $B$ is not fully contained in a single cube later).
I do not see how this generalize nicely to the general case $\mathcal{D}_{A}$ with $A \in \mathcal{B}_{\mathbf{R}^n}$ mentioned later in the proof. The same logic in the proof given would not have applied when showing the complement. Any suggestions?