# Uniqueness of Measures/Proof 1

## 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, for all $n \in \N$, $\mathcal{D}_n$ by:

$\mathcal{D}_n := \left\{{E \in \Sigma: \mu \left({G_n \cap E}\right) = \nu \left({G_n \cap E}\right)}\right\}$

Let us show that $\mathcal{D}_n$ is a Dynkin system.

By Intersection with Subset is Subset, $G_n \cap X = G_n$, whence $(3)$ implies that $X \in \mathcal{D}_n$.

Now, let $D \in \mathcal{D}_n$. Then:

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

Therefore, $X \setminus D \in \mathcal{D}_n$.

Finally, let $\left({D_m}\right)_{m \in \N}$ be a sequence of pairwise disjoint sets in $\mathcal{D}_n$.

Then:

 $\displaystyle \mu \left({G_n \cap \left({\bigcup_{m \mathop \in \N} D_m}\right)}\right)$ $=$ $\displaystyle \mu \left({\bigcup_{m \mathop \in \N} \left({G_n \cap D_m}\right)}\right)$ Intersection Distributes over Union $\displaystyle$ $=$ $\displaystyle \sum_{m \mathop \in \N} \mu \left({G_n \cap D_m}\right)$ $\mu$ is a measure $\displaystyle$ $=$ $\displaystyle \sum_{m \mathop \in \N} \nu \left({G_n \cap D_m}\right)$ $D_m \in \mathcal{D}_n$ $\displaystyle$ $=$ $\displaystyle \nu \left({G_n \cap \left({\bigcup_{m \mathop \in \N} D_m}\right)}\right)$ Above reasoning in opposite order

Therefore, $\displaystyle \bigcup_{m \mathop \in \N} D_m \in \mathcal{D}_n$.

Thus, we have shown that $\mathcal{D}_n$ is a Dynkin system.

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

$\forall n \in \N: \mathcal G \subseteq \mathcal{D}_n$
$\delta \left({\mathcal G}\right) = \sigma \left({\mathcal G}\right) = \Sigma$

where $\delta$ denotes generated Dynkin system.

By definition of $\delta \left({\mathcal G}\right)$, this means:

$\forall n \in \N: \delta \left({\mathcal G}\right) \subseteq \mathcal{D}_n$

That is, for all $n \in \N$, $\Sigma \subseteq \mathcal{D}_n \subseteq \Sigma$.

By definition of set equality:

$\Sigma = \mathcal{D}_n$ for all $n \in \N$

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

$\mu \left({G_n \cap E}\right) = \nu \left({G_n \cap E}\right)$

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

$\displaystyle \bigcup_{n \mathop \in \N} \left({G_n \cap E}\right) = \left({\bigcup_{n \mathop \in \N} G_n}\right) \cap E = X \cap E = E$

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

 $\displaystyle \mu \left({E}\right)$ $=$ $\displaystyle \lim_{n \to \infty} \mu \left({G_n \cap E}\right)$ Characterization of Measures: $(3)$ $\displaystyle$ $=$ $\displaystyle \lim_{n \to \infty} \nu \left({G_n \cap E}\right)$ $\forall n \in \N: E \in \mathcal{D}_n$ $\displaystyle$ $=$ $\displaystyle \nu \left({E}\right)$ Characterization of Measures: $(3)$

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

$\blacksquare$