Conditions for Equivalence of Measures Consider the following theorem for equivalence of measures:
Let $(E, \mathcal{A})$ be a measureable space with $\mathcal{A} = \sigma(\mathcal{G})$ for a ($\cap$-stable family $\mathcal{G}$) and  $\mu, \nu$ be measures on that spaces. If there exists a sequence $(G)_{n \in \mathbb{N}} \in \mathcal{G}$ which ascends to $E$ (ie. $G_n \uparrow E$) and $\mu(G_n), \nu(G_n) < \infty$ for all $n$ then
$$\mu(G) = \nu(G) ~~\forall G \in \mathcal{G} ~\implies~ \mu = \nu ~\text{on}~ \mathcal{A}.$$
Here are my questions:  


*

*Equivalently one can say, if ($\sigma$-finite) measures agree on $\mathcal{G}$,
then they are already equal on $\mathcal{A}$. How can I translate
this (more intuitive statement) to the "there exists a sequence"
statement?

*Why is $\mu(G_n), \nu(G_n) < \infty$ required? Can someone give a pathological example if this requirement is not fullfiled?
Thank you.

For (2), let $E = \mathbb{N}$ and $\mathcal{A}=2^{\mathbb{N}}$.  Let $\mathcal{G} = \{\{n, n+1, n+2, \dots\} : n \ge 0\}$.  Then $\mathcal{A} =\sigma(\mathcal{G})$ and $\mathcal{G}$ is stable under intersection.  There is a sequence in $\mathcal{G}$ ascending to $E$; in particular, $\mathbb{N} \in \mathcal{G}$ so we can take $G_n = \mathbb{N}$.  Let $\mu$ be counting measure and $\nu = 2\mu$.  It is true that $\mu(G) = \nu(G) = \infty$ for all $G \in \mathcal{G}$, but it is not true that $\mu=\nu$.
