Uniqueness of Measures/Proof 2

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:


 * $(3):\quad \forall G \in \mathcal G: \mu \left({G}\right) = \nu \left({G}\right)$
 * $(4):\quad \forall n \in \N: \mu \left({G_n}\right)$ is finite

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 the set:
 * $\mathcal{A} = \left\{{T \in \Sigma: \forall G \in \mathcal{G}: \mu \left({G \cap T}\right) = \nu \left({G \cap T}\right)}\right\}$

By Intersection with Subset is Subset, it follows that $G \cap X = G$ (for all $G \in \mathcal{G}$); therefore, $X \in \mathcal{A}$ by assumption $\left({3}\right)$.

Now, define the set:


 * $\Sigma' = \left\{{S \in \Sigma: \forall T \in \mathcal{A}: \mu \left({S \cap T}\right) = \nu \left({S \cap T}\right)}\right\}$

It follows directly from the definitions that $\mathcal{G} \subseteq \Sigma' \subseteq \Sigma$.

Note that because $X \in \mathcal{A}$ (from before), it follows that ${\mu \restriction_{\Sigma'}} = {\nu \restriction_{\Sigma'}}$ (where $\restriction$ denotes restriction).

By the definition of the $\sigma$-algebra generated by $\mathcal{G}$ and the equality of sets, it suffices to show that $\Sigma'$ is a $\sigma$-algebra over $X$.

This would then imply that $\Sigma = \Sigma'$, the desired result.


 * Proposition $1$. If $T \in \mathcal{A}$ and $G \in \mathcal{G}$, then $G \cap T \in \mathcal{A}$.

Proof. Let $H \in \mathcal{G}$ be arbitrary.

By the associativity of intersection, it follows that $H \cap \left({G \cap T}\right) = \left({H \cap G}\right) \cap T$.

The result follows because $H \cap G \in \mathcal{G}$ by assumption $\left({1}\right)$.


 * Proposition $2$. If $T \in \mathcal{A}$ and $S \in \Sigma'$, then $S \cap T \in \mathcal{A}$.

Proof. Let $G \in \mathcal{G}$ be arbitrary.

By the associativity and commutativity of intersection, $G \cap \left({S \cap T}\right) = S \cap \left({G \cap T}\right)$.

The result follows by the definition of $\Sigma'$, and because $G \cap T \in \mathcal{A}$ by Proposition $1$.


 * Lemma $1$. If $A, B \in \Sigma'$, then $A \cap B \in \Sigma'$.

Proof. Let $T \in \mathcal{A}$ be arbitrary.

By the associativity of intersection, it follows that $\left({A \cap B}\right) \cap T = A \cap \left({B \cap T}\right)$.

The result follows because $B \cap T \in \mathcal{A}$ by Proposition $2$.


 * Proposition $3$. If $A, B \in \Sigma'$ and $\mu \left({A}\right)$ is finite, then $A \cup B \in \Sigma'$.

Proof. Let $T \in \mathcal{A}$ be arbitrary. Then:

and similarly for $\nu$.

The result follows because $A \cap B \in \Sigma'$ by Lemma $1$.


 * Corollary $1$.
 * $\displaystyle \forall n \in \N: \bigcup_{k \mathop = 0}^n G_k \in \Sigma'$

Proof. The result follows by assumption $\left({4}\right)$, Proposition $3$, the associativity of intersection, and mathematical induction on $n$.


 * Lemma $2$. If $S \in \Sigma'$, then $X \setminus S \in \Sigma'$.

Proof. Let $T \in \mathcal{A}$ be arbitrary. Then:

and similarly for $\nu$.

The result follows by Corollary $1$ and Lemma $1$.


 * Corollary $2$. $\Sigma'$ is an algebra of sets over $X$.

Proof. The result follows from Lemmas $1$ and $2$, and De Morgan's laws.


 * Final Step. $\Sigma'$ is a $\sigma$-algebra over $X$.

Proof. By Corollary $2$, it suffices to show that $\Sigma'$ is a $\sigma$-ring.

Let $\left({S_k}\right)_{k \mathop = 0}^{\infty}$ be a sequence of sets in $\Sigma'$, and let $T \in \mathcal{A}$ be arbitrary.

By the distributivity of intersection over union and Characterization of Measures: $\left({3}\right)$, it follows that:
 * $\displaystyle \mu \left({\bigcup_{k \mathop = 0}^{\infty} S_k \cap T}\right) = \lim_{n \to \infty} \mu \left({\bigcup_{k \mathop = 0}^n S_k \cap T}\right)$

and similarly for $\nu$.

By Corollary $2$, it follows by the associativity of intersection and by mathematical induction on $n$ that:
 * $\displaystyle \forall n \in \N: \bigcup_{k \mathop = 0}^n S_k \in \Sigma'$

Therefore, $\Sigma'$ is a $\sigma$-ring, and the theorem is proven.