Proving two measures of Borel sigma-algebra are equal

By Dynkin's $\pi$-$\lambda$ Theorem two finite measures of the same mass that agree on a $\pi$-system agree on the $\sigma$-algebra generated by that $\pi$-system.

To prove this using Dynkin, consider $\Lambda:=\{ B \in \mathcal{B} : \mu_1(B \cap (a,b)) = \mu_2(B \cap (a,b))\}$ is a Dynkin system containing the $\pi$-system of open intervals. Thus $\Lambda$ contains the $\sigma$-algebra generated by open intervals (which is $\mathcal{B}$) and hence $\Lambda = \mathcal{B}$. Thus the restriction of $\mu_1$ and the restriction of $\mu_2$ are identical on any finite interval. An arbitrary set $B$ can be written as the increasing union of $B\cap (-n,n)$ for larger and larger $n$, so the result follows.