The statement says that if $\mu'$ is a measure defined on the $\sigma$-algebra $\Sigma$ of all the Lebesgue measurable subsets of $\mathbb{R}$, s.t. $\mu' \left(a, b \right) = b-a$ for any open interval $ \left(a, b \right)$, then it is indeed Lebesgue measure $\mu$ (which is induced by the Lebesgue outer measure).
I have already shown that for all the open $U\in \Sigma$, $\mu'\left(U\right)=\mu\left(U\right)$. But how to show that $\forall A \in \Sigma$, $\mu'\left(A\right) = \mu \left(A\right)$ in general?
I think there might be a general theorem says that if two measures give the same value on all open sets, then these two measures are the same. Is it true or false?