Uniqueness of Measures/Proof 1

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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, 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:

\(\ds \map \mu {G_n \cap \paren {X \setminus D} }\) \(=\) \(\ds \map \mu {G_n \setminus D}\) Intersection with Set Difference is Set Difference with Intersection
\(\ds \) \(=\) \(\ds \map \mu {G_n} - \map \mu {G_n \cap D}\) $\map \mu {G_n} < +\infty$, Set Difference and Intersection form Partition, Measure is Finitely Additive Function
\(\ds \) \(=\) \(\ds \map \nu {G_n} - \map \nu {G_n \cap D}\) $(3)$, $D \in \DD_n$
\(\ds \) \(=\) \(\ds \map \nu {G_n \cap \paren {X \setminus D} }\) Above reasoning in opposite order

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:

\(\ds \map \mu {G_n \cap \paren {\bigcup_{m \mathop \in \N} D_m} }\) \(=\) \(\ds \map \mu {\bigcup_{m \mathop \in \N} \paren {G_n \cap D_m} }\) Intersection Distributes over Union
\(\ds \) \(=\) \(\ds \sum_{m \mathop \in \N} \map \mu {G_n \cap D_m}\) $\mu$ is a measure
\(\ds \) \(=\) \(\ds \sum_{m \mathop \in \N} \map \nu {G_n \cap D_m}\) $D_m \in \DD_n$
\(\ds \) \(=\) \(\ds \map \nu {G_n \cap \paren {\bigcup_{m \mathop \in \N} D_m} }\) Above reasoning in opposite order

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:

$\forall n \in \N: \Sigma \subseteq \DD_n \subseteq \Sigma$

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 $E \in \Sigma$:

\(\ds \map \mu E\) \(=\) \(\ds \lim_{n \mathop \to \infty} \map \mu {G_n \cap E}\) Characterization of Measures: $(3)$
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \map \nu {G_n \cap E}\) $\forall n \mathop \in \N: E \in \DD_n$
\(\ds \) \(=\) \(\ds \map \nu E\) Characterization of Measures: $(3)$

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

$\blacksquare$


Sources

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