Uniqueness of Measures/Proof 2

From ProofWiki
Jump to navigation Jump to search

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 $\left({1}\right)$.

$\Box$


  • 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 $1$.

$\Box$


  • 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 $2$.

$\Box$


  • 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:

\(\displaystyle \mu \left({\left({A \cup B}\right) \cap T}\right)\) \(=\) \(\displaystyle \mu \left({\left({A \cap T}\right) \cup \left({B \cap T}\right)}\right)\) by the commutativity of intersection and the distributivity of intersection over union
\(\displaystyle \) \(=\) \(\displaystyle \mu \left({A \cap T}\right) + \mu \left({B \cap T}\right) - \mu \left({\left({A \cap T}\right) \cap \left({B \cap T}\right)}\right)\) because a measure is strongly additive; this expression is defined (in the extended real numbers $\overline{\R}$) because of Intersection is Subset and the monotonicity of $\mu$ (recall that $\mu \left({A}\right)$ is finite by assumption)
\(\displaystyle \) \(=\) \(\displaystyle \mu \left({A \cap T}\right) + \mu \left({B \cap T}\right) - \mu \left({\left({A \cap B}\right) \cap T}\right)\) 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 $\left({4}\right)$, 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 $T \in \mathcal{A}$ be arbitrary. Then:

\(\displaystyle \mu \left({\left({X \setminus S}\right) \cap T}\right)\) \(=\) \(\displaystyle \lim_{n \to \infty} \mu \left({\bigcup_{k \mathop = 0}^n G_k \cap T \setminus S}\right)\) by assumption $\left({2}\right)$, by Intersection with Set Difference is Set Difference with Intersection, the commutativity of intersection, the distributivity of intersection over union, the right distributivity of set difference over union, and by Characterization of Measures: $\left({3}\right)$
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \to \infty} \left({\mu \left({\bigcup_{k \mathop = 0}^n G_k \cap T}\right) - \mu \left({\bigcup_{k \mathop = 0}^n G_k \cap S \cap T}\right)}\right)\) by Set Difference and Intersection form Partition, the additivity of $\mu$, and the associativity and commutativity of intersection; this expression is defined because of Intersection is Subset, the monotonicity of $\mu$, and the subadditivity of $\mu$, by assumption $\left({4}\right)$

and similarly for $\nu$.

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

$\Box$


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

$\Box$


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.

$\blacksquare$