Question: 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

Answer: 
The collection of $A \in \mathcal{E}$ such that $\mu(A \cap X\_h) = \nu(A \cap X\_h)$ for all $h$ forms a $\sigma$-algebra containing $\sigma(G)$.
Note that if $\nu$ is a $\sigma$-finite non negative measure then on a $\sigma$-algebra $G$ then every finite nonnegative measurable function $f$ defines the $\sigma$-finite measure $\mu:=f.\nu$. We can take $f$ to be a characteristic function. This is because, if $f$ is integrable then it defines a set function $\mu(A)= \int\_A f d\nu$.

In general, statements involving $\sigma$-finite "situations" can always be reduced to the finite case (see Bogachev section 2.6).