# Uniqueness of Measures/Proof 2

## Theorem

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

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 $(1)$.

$\Box$

**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 $1$.

$\Box$

**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 $2$.

$\Box$

**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:

\(\ds \map \mu {\paren {A \cup B} \cap T}\) | \(=\) | \(\ds \map \mu {\paren {A \cap T} \cup \paren {B \cap T} }\) | Intersection is Commutative and Intersection Distributes over Union | |||||||||||

\(\ds \) | \(=\) | \(\ds \map \mu {A \cap T} + \map \mu {B \cap T} - \map \mu {\paren {A \cap T} \cap \paren {B \cap T} }\) | Measure is Strongly Additive | |||||||||||

\(\ds \) | \(=\) | \(\ds \map \mu {A \cap T} + \map \mu {B \cap T} - \map \mu {\paren {A \cap B} \cap T}\) | Intersection is Associative, Intersection is Commutative and Intersection is Idempotent |

and similarly for $\nu$.

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

$\Box$

**Corollary $1$.**

- $\displaystyle \forall n \in \N: \bigcup_{k \mathop = 0}^n G_k \in \Sigma'$

*Proof.* The result follows by assumption $(4)$, Proposition $3$, the associativity of intersection, and mathematical induction on $n$.

$\Box$

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

*Proof.* Let $T \in \AA$ be arbitrary. Then:

\(\ds \map \mu {\paren {X \setminus S} \cap T}\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \map \mu {\bigcup_{k \mathop = 0}^n G_k \cap T \setminus S}\) | 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: $(3)$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\map \mu {\bigcup_{k \mathop = 0}^n G_k \cap T} - \map \mu {\bigcup_{k \mathop = 0}^n G_k \cap S \cap T} }\) | Set Difference and Intersection form Partition, 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 $(4)$ |

and similarly for $\nu$.

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

$\Box$

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

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

$\Box$

**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.

$\blacksquare$