I am having difficulty proving that the lebesgue measure $\lambda$ in the question below is the only one that satisfies the productory property. I imagine I need to consider an arbitrary measure $\mu$ in this measurable space and show that $\mu(A)=\lambda(A)$ with $A \in B(\mathbb{R}^d)$.
The question also says to consider $B(\mathbb{R}^d)$ generated by the squares in the statement $(Q)$ and, in fact, I observed that a collection of closed sets in $\mathbb{R}^d $ generate $B(\mathbb{R}^d)$. I first thought about equating $\mu(A)=\lambda(A)$ with the product, but that seems too simple, I also thought about showing that the class $V=\{A \in B(\mathbb{R}^d)| \mu(A)=\lambda(A)\}$ is sigma-algebra, but I'm not sure. What can I do here?
Question: Show that the Lebesgue measure is the only measure on $(\mathbb{R}^d, B(\mathbb{R}^d))$ such that $λ(Q)=\prod_{i=1}^d(b_i−a_i)$, where $Q=\prod_{i=1}^d [a_i, b_i]$ for $a, b \in \mathbb{R}^d$ such that $a_i \leq b_i$ for $i = 1, 2, \ldots , d$. You can use that $B(\mathbb{R}^d)$ is generated by the squares as above.