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: {{begin-eqn}} {{eqn|l = \mu \left({G_n \cap \left({X \setminus D}\right)}\right)      |r = \mu \left({G_n \setminus D}\right)      |c = Intersection with Set Difference is Set Difference with Intersection }} {{eqn|r = \mu \left({G_n}\right) - \mu \left({G_n \cap D}\right)      |c = $\mu \left({G_n}\right) < +\infty$, Set Difference and Intersection form Partition, Measure is Finitely Additive Function }} {{eqn|r = \nu \left({G_n}\right) - \nu \left({G_n \cap D}\right)      |c = $(3)$, $D \in \mathcal{D}_n$ }} {{eqn|r = \nu \left({G_n \cap \left({X \setminus D}\right)}\right)      |c = Above reasoning in opposite order }} {{end-eqn}} 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: {{begin-eqn}} {{eqn|l = \mu \left({G_n \cap \left({\bigcup_{m \mathop \in \N} D_m}\right)}\right)      |r = \mu \left({\bigcup_{m \mathop \in \N} \left({G_n \cap D_m}\right)}\right)      |c = Intersection Distributes over Union }} {{eqn|r = \sum_{m \mathop \in \N} \mu \left({G_n \cap D_m}\right)      |c = $\mu$ is a measure }} {{eqn|r = \sum_{m \mathop \in \N} \nu \left({G_n \cap D_m}\right)      |c = $D_m \in \mathcal{D}_n$ }} {{eqn|r = \nu \left({G_n \cap \left({\bigcup_{m \mathop \in \N} D_m}\right)}\right)      |c = Above reasoning in opposite order }} {{end-eqn}} 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: :$\forall n \in \N: \mathcal G \subseteq \mathcal{D}_n$ From $(1)$ and Dynkin System with Generator Closed under Intersection is Sigma-Algebra: :$\delta \left({\mathcal G}\right) = \sigma \left({\mathcal G}\right) = \Sigma$ 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: :$\displaystyle \bigcup_{n \mathop \in \N} \left({G_n \cap E}\right) = \left({\bigcup_{n \mathop \in \N} G_n}\right) \cap E = X \cap E = E$ from Intersection Distributes over Union and Intersection with Subset is Subset. Thus, for all $E \in \Sigma$: {{begin-eqn}} {{eqn|l = \mu \left({E}\right)      |r = \lim_{n \to \infty} \mu \left({G_n \cap E}\right)      |c = Characterization of Measures: $(3)$ }} {{eqn|r = \lim_{n \to \infty} \nu \left({G_n \cap E}\right)      |c = $\forall n \in \N: E \in \mathcal{D}_n$ }} {{eqn|r = \nu \left({E}\right)      |c = Characterization of Measures: $(3)$ }} {{end-eqn}} That is to say, $\mu = \nu$. {{qed}}