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

Proof
Define, for all $n \in \N$, $\mathcal{D}_n$ by:


 * $\mathcal{D}_n := \left\{{E \in \Sigma: \mu \left({G_n \cap E}\right) = \nu \left({G_n \cap E}\right)}\right\}$

Let us show that $\mathcal{D}_n$ is a Dynkin system.

By Intersection with Subset is Subset, $G_n \cap X = G_n$, whence $(3)$ implies that $X \in \mathcal{D}_n$.

Now, let $D \in \mathcal{D}_n$. Then:

Therefore, $X \setminus D \in \mathcal{D}_n$.

Finally, let $\left({D_m}\right)_{m \in \N}$ be a sequence of pairwise disjoint sets in $\mathcal{D}_n$.

Then:

Therefore, $\displaystyle \bigcup_{m \mathop \in \N} D_m \in \mathcal{D}_n$.

Thus, we have shown that $\mathcal{D}_n$ is a Dynkin system.

Combining $(1)$ and $(3)$, it follows that:


 * $\forall n \in \N: \mathcal G \subseteq \mathcal{D}_n$

From $(1)$ and Dynkin System with Generator Closed under Intersection is Sigma-Algebra:


 * $\delta \left({\mathcal G}\right) = \sigma \left({\mathcal G}\right) = \Sigma$

where $\delta$ denotes generated Dynkin system.

By definition of $\delta \left({\mathcal G}\right)$, this means:


 * $\forall n \in \N: \delta \left({\mathcal G}\right) \subseteq \mathcal{D}_n$

That is, for all $n \in \N$, $\Sigma \subseteq \mathcal{D}_n \subseteq \Sigma$.

Hence by Equality of Sets, $\Sigma = \mathcal{D}_n$ for all $n \in \N$.

Thus, for all $n \in \N$ and $E \in \Sigma$:


 * $\mu \left({G_n \cap E}\right) = \nu \left({G_n \cap E}\right)$

Now, from Intersection Preserves Subsets, $E_n := G_n \cap E$ defines an increasing sequence of sets with limit:


 * $\displaystyle \bigcup_{n \mathop \in \N} \left({G_n \cap E}\right) = \left({\bigcup_{n \mathop \in \N} G_n}\right) \cap E = X \cap E = E$

from Intersection Distributes over Union and Intersection with Subset is Subset.

Thus, for all $E \in \Sigma$:

That is to say, $\mu = \nu$.