Uniqueness of Measures/Proof 1

Theorem
Let $\left({X, \Sigma}\right)$ be a measurable space.

Let $\mathcal G \subseteq \mathcal P \left({X}\right)$ be a generator for $\Sigma$; i.e., $\Sigma = \sigma \left({\mathcal G}\right)$.

Suppose that $\mathcal G$ satisfies the following conditions:


 * $(1):\quad \forall G, H \in \mathcal G: G \cap H \in \mathcal G$
 * $(2):\quad$ There exists an exhausting sequence $\left({G_n}\right)_{n \in \N} \uparrow X$ in $\mathcal G$

Let $\mu, \nu$ be measures on $\left({X, \mathcal A}\right)$, and suppose that:


 * $(3):\quad \forall G \in \mathcal G: \mu \left({G}\right) = \nu \left({G}\right)$
 * $(4):\quad \forall n \in \N: \mu \left({G_n}\right), \nu \left({G_n}\right) < +\infty$

Then $\mu = \nu$.

Alternatively, the exhausting sequence in $(2)$ may be replaced by a countable $\mathcal G$-cover $\left({G_n}\right)_{n \in \N}$, still subject to $(4)$.