Uniqueness of Measures/Proof 2

== Theorem ==
== 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, $(3)$:
:$X \in \AA$
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 :
:$X \in \AA$
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 $(1)$.
* 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 $1$.
* 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 $2$.
* 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:
and similarly for $\nu$.
The result follows because $A \cap B \in \Sigma'$ by Lemma $1$.
* 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 $n$.
* Lemma $2$. If $S \in \Sigma'$, then $X \setminus S \in \Sigma'$.
Proof. Let $T \in \AA$ 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 $\sequence {S_k}_{k \mathop = 0}^\infty$ 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.
Category:Uniqueness of Measures