Lebesgue measure compact symmetric intervals Let $\mu$  be a measure on $L_m$ , ($m \ge 1$) - $\sigma$ - algebra of Lebesgue-measurable sets, such that its values on compact, symmetric intervals (cubes) are equal to Lebesgue measure of those cubes. Then $\mu = \mathcal{L}^m$.
Could you tell me how to prove this observation?
I know we can define Lebesgue measure by symmetric cubes, but I am not sure how to use it here.
Thank you.

This fact is false. Let $g(x)=1+\frac{1}{1+x^2}$ for $x\ge 0$ and $g(x)=1-\frac{1}{1+x^2}$ otherwise. Let define $\mu$ in $R$ as follows: $\mu(A)=\int_{A}g(x)d \cal{L}^1(x)$ for all Lebesgue measurable subset $A \subseteq R$. It is obvious that $\mu$  satisfies all conditions above but $\mu \neq \cal{L}^1$.
