Theorem: Uniqueness of Measures

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\GG \subseteq \powerset X$ be a generator for $\Sigma$; that is, $\Sigma = \map \sigma \GG$.
Suppose that $\GG$ satisfies the following conditions:
:$(1):\quad \forall G, H \in \GG: G \cap H \in \GG$
:$(2):\quad$ There exists an exhausting sequence $\sequence {G_n}_{n \mathop \in \N} \uparrow X$ in $\GG$
Let $\mu, \nu$ be measures on $\struct {X, \Sigma}$, and suppose that:
:$(3):\quad \forall G \in \GG: \map \mu G = \map \nu G$
:$(4):\quad \forall n \in \N: \map \mu {G_n}$ is finite
Then:
:$\mu = \nu$

Proof:
Define the set:
:$\AA = \set {T \in \Sigma: \forall G \in \GG: \map \mu {G \cap T} = \map \nu {G \cap T} }$
By Intersection with Subset is Subset, it follows that:
:$\forall G \in \GG: G \cap X = G$
Therefore, {{hypothesis}} $(3)$:
:$X \in \AA$
Now, define the set:
:$\Sigma' = \set {S \in \Sigma: \forall T \in \AA: \map \mu {S \cap T} = \map \nu {S \cap T} }$
It follows directly from the definitions that:
:$\GG \subseteq \Sigma' \subseteq \Sigma$
We have {{apriori}}:
:$X \in \AA$
Hence:
:$\mu {\restriction_{\Sigma'} } = \nu {\restriction_{\Sigma'} }$
where $\restriction$ denotes restriction.
By the definition of the $\sigma$-algebra generated by $\GG$ and set equality, 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 $T \in \AA$ and $G \in \GG$, then $G \cap T \in \AA$.
''Proof.'' Let $H \in \GG$ be arbitrary.
By the associativity of intersection, it follows that $H \cap \paren {G \cap T} = \paren {H \cap G} \cap T$.
The result follows because $H \cap G \in \GG$ by assumption $(1)$.
{{qed|lemma}}
* '''Proposition $2$.''' If $T \in \AA$ and $S \in \Sigma'$, then $S \cap T \in \AA$.
''Proof.'' Let $G \in \GG$ be arbitrary.
By the associativity and commutativity of intersection, $G \cap \paren {S \cap T} = S \cap \paren {G \cap T}$.
The result follows by the definition of $\Sigma'$, and because $G \cap T \in \AA$ by Proposition $1$.
{{qed|lemma}}
* '''Lemma $1$.''' If $A, B \in \Sigma'$, then $A \cap B \in \Sigma'$.
''Proof.'' Let $T \in \AA$ be arbitrary.
By the associativity of intersection, it follows that $\paren {A \cap B} \cap T = A \cap \paren {B \cap T}$.
The result follows because $B \cap T \in \AA$ by Proposition $2$.
{{qed|lemma}}
* '''Proposition $3$.''' If $A, B \in \Sigma'$ and $\map \mu A$ is finite, then $A \cup B \in \Sigma'$.
''Proof.'' Let $T \in \AA$ be arbitrary. Then:
{{begin-eqn}}
{{eqn | l = \map \mu {\paren {A \cup B} \cap T}
      | r = \map \mu {\paren {A \cap T} \cup \paren {B \cap T} }
      | c = Intersection is Commutative and Intersection Distributes over Union
}}
{{eqn | r = \map \mu {A \cap T} + \map \mu {B \cap T} - \map \mu {\paren {A \cap T} \cap \paren {B \cap T} }
      | c = Measure is Strongly Additive
}}
{{eqn | r = \map \mu {A \cap T} + \map \mu {B \cap T} - \map \mu {\paren {A \cap B} \cap T}
      | c = Intersection is Associative, Intersection is Commutative and Set Intersection is Idempotent
}}
{{end-eqn}}
and similarly for $\nu$.
The result follows because $A \cap B \in \Sigma'$ by Lemma $1$.
{{qed|lemma}}
* '''Corollary $1$.'''
:$\ds \forall n \in \N: \bigcup_{k \mathop = 0}^n G_k \in \Sigma'$
''Proof.'' The result follows by assumption $(4)$, Proposition $3$, the associativity of intersection, and mathematical induction on $n$.
{{qed|lemma}}
* '''Lemma $2$.''' If $S \in \Sigma'$, then $X \setminus S \in \Sigma'$.
''Proof.'' Let $T \in \AA$ be arbitrary. Then:
{{begin-eqn}}
{{eqn | l = \map \mu {\paren {X \setminus S} \cap T}
      | r = \lim_{n \mathop \to \infty} \map \mu {\bigcup_{k \mathop = 0}^n G_k \cap T \setminus S}
      | c = by assumption $(2)$, Intersection with Set Difference is Set Difference with Intersection, Intersection is Commutative, Intersection Distributes over Union, Set Difference is Right Distributive over Union, and by Characterization of Measures: $(3)$ <!-- maybe we want (yet) another version of "Characterization of Measures: $(3)$" specifically for countable unions? -->
}}
{{eqn | r = \lim_{n \mathop \to \infty} \paren {\map \mu {\bigcup_{k \mathop = 0}^n G_k \cap T} - \map \mu {\bigcup_{k \mathop = 0}^n G_k \cap S \cap T} }
      | c = Set Difference and Intersection form Partition<!-- should we (or do we already) have a page for the case where one or more of the sets in question are allowed to be empty? -->, Measure is Finitely Additive Function, Intersection is Associative, Intersection is Commutative; this expression is defined because of Intersection is Subset, the monotonicity of $\mu$, and the subadditivity of $\mu$, by assumption $(4)$
}}
{{end-eqn}}
and similarly for $\nu$.
The result follows by Corollary $1$ and Lemma $1$.
{{qed|lemma}}
* '''Corollary $2$.''' $\Sigma'$ is an algebra of sets over $X$.
''Proof.'' The result follows from Lemmas $1$ and $2$, and De Morgan's laws.
<!-- separate page for a detailed explanation of the above? -->
{{qed|lemma}}
* '''Final Step.''' $\Sigma'$ is a $\sigma$-algebra over $X$.
''Proof.'' By Corollary $2$, it suffices to show that $\Sigma'$ is a $\sigma$-ring.
Let $\sequence {S_k}_{k \mathop = 0}^\infty$ be a sequence of sets in $\Sigma'$, and let $T \in \AA$ be arbitrary.
By the Intersection Distributes over Union and Characterization of Measures: $(3)$:
:$\ds \map \mu {\bigcup_{k \mathop = 0}^\infty S_k \cap T} = \lim_{n \mathop \to \infty} \map \mu {\bigcup_{k \mathop = 0}^n S_k \cap T}$
and similarly for $\nu$.
By Corollary $2$, it follows by the associativity of intersection and by mathematical induction on $n$ that:
:$\ds \forall n \in \N: \bigcup_{k \mathop = 0}^n S_k \in \Sigma'$
Therefore, $\Sigma'$ is a $\sigma$-ring, and the theorem is proven.
{{qed}}