The following theorem has been mentioned (and partially proved) in the book Functions of Bounded Variation and Free Discontinuity Problems by Luigi Ambrosio et. al.
Let $\mu,\nu$ be positive measures on $(X,\mathcal{E})$. Assume that they are equal on a collection of sets $\mathcal{G}$ which is closed under finite intersections. Also assume that there are sets $X_h \in \mathcal{G}$ such that $\displaystyle{X= \bigcup_{h=1}^\infty X_h}$ and that $\mu(X_h) = \nu(X_h) < \infty$ for all $h$. Then $\mu, \nu$ are equal on the $\sigma$-algebra generated by $\mathcal{G}$.
The authors prove the theorem for the case where $\mu,\nu$ are positive finite measures and $\mu(X) = \nu(X)$ and say that the general case follows easily. However, this is not at all straight forward for me. I have tried to prove this but cannot reach a valid proof. Here is my attempt at a proof:
Let $G_h = \{g \cap X_h | g \in G\}$. Then clearly from using the finite case of the theorem, we have that $\mu,\nu$ coincide on every $\sigma (G_h)$. My problem now is to show that this implies that they agree on $\sigma(G)$. All my attempts in this direction have been futile.
I feel that the solution should be relatively easy (as the authors themselves point out). Any help in the proof is greatly appreciated.
Thanks, Phanindra