Uniqueness of Measures/Proof 2

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 $G \cap X = G$ (for all $G \in \GG$); therefore, $X \in \AA$ $(3)$.

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 $\GG \subseteq \Sigma' \subseteq \Sigma$.

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

By the definition of the $\sigma$-algebra generated by $\GG$ 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 $T \in \AA$ 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.