Is the Lebesgue measure on the Borel $\sigma$-algebra the only measure which is equal to the volume of a closed rectangle on each closed rectangle?

Question

Let $R_d$ be a d-dimensional closed rectangle. I want to show that each measure $\mu$ defined on the Borel $\sigma$-algebra for which \begin{equation} \mu(R_d)=l(R_d) \qquad \forall R_d \end{equation} is equal to the restriction of the Lebesgue measure on $\mathcal{B}^d$: $\mu\equiv m$. Where $l(R_d)$ is the volume of the rectangle. I know that an outer measure $\mu:\mathcal{B}^d\to[0,\infty]$ has the following properties:

1. $\mu \cdot(\emptyset)=0$

2. $A\subset B \Rightarrow \mu \cdot (A)\leq\mu\cdot (B)$

3. $A_n\subset\mathcal{B}^d, n\in\mathbb{N} \Rightarrow \mu(\bigcup\limits_{n=1}^{\infty} A_{n})\leq \sum_{n=1}^{\infty}\mu(A_n)$

In addition, every closed rectangle is $m$-measurable, thus \begin{equation} m(Z)=m(Z \cap R_d)+m(Z\cap R_d^C), \end{equation} and that $m(R_d)=l(R_d)$.

I am very lost on how to proceed and would be very grateful for any help. Thank you!