Considering the following Theorem:
If $(X,\mathcal{A})$ a measurable space and $\mathcal{A}=\sigma(\mathcal{G})$ such that
- $\mathcal{G}$ is stable under finite intersections
- there exists and exahusting sequence $(G_j)_{j \in \mathbb{N}} \subset\mathcal{G}$ with $G_j \uparrow X$
Then, any two measures $\mu,\nu$ that coincide on $\mathcal{G}$ and are finite for all members of $(G_j)_{j \in \mathbb{N}}$ are equal, i.e. $\mu(A) = \nu(A)$, $\forall A \in \mathcal{A}$.
I'm trying to show that the Theorem still holds for any $(G_j)_{j \in \mathbb{N}}\subset\mathcal{G}$ such that $\bigcup\limits_{j \in \mathbb{N}} G_j = X$ and $\nu(G_j) = \mu(G_j) < \infty$
What I've done so far is
- Define $F_k = \bigcup\limits_{j=1}^{k} G_j$, which is an exhausting sequence such that $\bigcup\limits_{j=1}^{\infty} F_j = \bigcup\limits_{j=1}^{\infty} G_j = X$
- Shown by induction that $\nu(F_n),\mu(F_n)<\infty$
Now I'm struggling to show that $\nu(F_n) = \mu(F_n)$.
I tried to do this by induction as well and got to the following expresion
$$\nu(F_{n+1}) = \mu(F_{n+1}) \iff \nu\left(\bigcup\limits_{j=1}^{n} G_j \cap G_{n+1}\right) = \mu\left(\bigcup\limits_{j=1}^{n} G_j \cap G_{n+1}\right)$$
But I'm not sure what to do next.
Am I in the right track or should I consider to show this last part somehow else?
Any suggestions are highly appreciated!
Claim: $\mu (\bigcup_{i=1}^{n} A_i)=\nu (\bigcup_{i=1}^{n} A_i)$ for any finite collection $\{A_1,A_2,...,A_n\}$ in $\mathcal G$.
Proof by induction on $n$:
$n=1$ No problem.
$n=2$ Use the fact that $\mu (A_1\cup A_2)=\mu(A_1)+\mu(A_2)-\mu(A_1 \cap A_2)$ and a similar equation for $\nu$.
Suppose the result holds for $n$ and consider $n+1$ sets $\{A_1,A_2,...,A_n,A_{n+1}\}$ in $\mathcal G$.
Use the equation $\mu (A\cup B)=\mu(A)+\mu(B)-\mu(A \cap B)$ with $A=A_{n+1}$ and $B=\bigcup_{i=1}^{n}A_i$