Uniqueness of Measures/Proof 1

Proof
Define, for all $n \in \N$, $\DD_n$ by:


 * $\DD_n := \set {E \in \Sigma: \map \mu {G_n \cap E} = \map \nu {G_n \cap E} }$

Let us show that $\DD_n$ is a Dynkin system.

By Intersection with Subset is Subset, $G_n \cap X = G_n$, whence $(3)$ implies that $X \in \DD_n$.

Now, let $D \in \DD_n$. Then:

Therefore, $X \setminus D \in \DD_n$.

Finally, let $\sequence {D_m}_{m \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\DD_n$.

Then:

Therefore:
 * $\ds \bigcup_{m \mathop \in \N} D_m \in \DD_n$.

Thus, we have shown that $\DD_n$ is a Dynkin system.

Combining $(1)$ and $(3)$, it follows that:


 * $\forall n \in \N: \GG \subseteq \DD_n$

From $(1)$ and Dynkin System with Generator Closed under Intersection is Sigma-Algebra:


 * $\map \delta \GG = \map \sigma \GG = \Sigma$

where $\delta$ denotes generated Dynkin system.

By definition of $\map \delta \GG$, this means:


 * $\forall n \in \N: \map \delta \GG \subseteq \DD_n$

That is, for all $n \in \N$, $\Sigma \subseteq \DD_n \subseteq \Sigma$.

By definition of set equality:
 * $\Sigma = \DD_n$ for all $n \in \N$

Thus, for all $n \in \N$ and $E \in \Sigma$:


 * $\map \mu {G_n \cap E} = \map \nu {G_n \cap E}$

Now, from Set Intersection Preserves Subsets, $E_n := G_n \cap E$ defines an increasing sequence of sets with limit:


 * $\ds \bigcup_{n \mathop \in \N} \paren {G_n \cap E} = \paren {\bigcup_{n \mathop \in \N} G_n} \cap E = X \cap E = E$

from Intersection Distributes over Union and Intersection with Subset is Subset.

Thus, for all $E \in \Sigma$:

That is to say, $\mu = \nu$.