What i have to prove:
For X and $\mathcal{F}$ not empty let $( X, \mathcal{A})$ be a measureable space with $\mathcal{A} = \sigma(\mathcal{F})$ for a closed under intersection family $\mathcal{F}$ and $\mu, \nu$ measures:$\sigma(\mathcal{F})$->[0,+$\infty$] , where $\mathcal{F} \subseteq \mathcal{P(}X)$ and $\mu(F)=\nu(F)$ for every F$\in \mathcal{F}$. If there exists disjoint F1,F2,F3,…..$\in \mathcal{F}$ such that X= $\bigcup_{n=1}^\infty Fn$ and $\mu(F_n)= \nu(F_n) < \infty$ for all $n$ then
$$\mu(G) = \nu(G) ~~\forall G \in \mathcal{A}=
\sigma(\mathcal{F})$$
And i know the following theorem for equivalence of measures $\mu(A)=\nu(A)$ for every A $\in \sigma(\mathcal{F})$:
Let $( X, \mathcal{A})$ be a measureable space with $\mathcal{A} = \sigma(\mathcal{F})$ for a closed under $\cap$– family $\mathcal{F}$ and $\mu, \nu$ be measures on that spaces, where $\mathcal{F} \subseteq \mathcal{P(}X)$ and $\mu(F)=\nu(F)$ for every F$\in \mathcal{F}$. If there exists an increasing sequence $(Fn)_{n \in \mathbb{N}} \in \mathcal{F}$ such that X=the countable union from n=1 to infinity of Fn and $\mu(F_n)= \nu(F_n) < \infty$ for all $n$ then
$$\mu(G) = \nu(G) ~~\forall G \in \mathcal{A}=
\sigma(\mathcal{F})$$
a first thought of mine would be to create an increasing sequence with the disjoint sets but ultimately there is nothing that tells us that the term Bn for instance will have finite measure. So, i can't think of an increasing sequence which doesn't do this(unless we can somehow show that it converges and therefore is bounded, could it have to do with taking some sort of intersection of the disjoint sets(?))
I'm starting to think that i have to prove it without creating an increasing sequence and it's probably something more general.I thought of proving that the set in which $\mu(G)=\nu(G)$ is in fact $\sigma(\mathcal{F})$, by showing that it's Dynkin but i think i need the increasing sequence there as well.
Any hints? I would like it if someone could give me a hint perhaps in the comments and of course post the answer for others to see, should they want to do so.
Best Answer
For $n\in \mathbb N,$ let $\mathcal C_n = \{G\in \sigma(\mathcal F) : \mu(G\cap F_n) = \nu(G\cap F_n)\}.$ If we can show each $\mathcal C_n = \sigma(\mathcal F),$ we're done, since then, for any $G\in \mathcal \sigma(F),$ $\mu(G) = \sum_n \mu(G\cap F_n) = \sum_n \nu(G\cap F_n) = \nu(G).$
We have $\mathcal F\subseteq \mathcal C_n,$ so it suffices to show $\mathcal C_n$ is a $\sigma$-algebra. By Dynkin's theorem, it suffices to show that $\mathcal C_n$ contains $X$ and is closed under countable increasing unions and contained differences (i.e. $B\setminus A$ for $A\subseteq B).$ That it contains $X$ is clear, and closure under increasing unions follows from continuity of measure. For contained differences, if $A\subseteq B$ and $A,B\in \mathcal C_n,$ then since $\mu(F_n)=\nu(F_n)<\infty,$ $$\mu((B\setminus A)\cap F_n) = \mu(B\cap F_n)- \mu(A\cap F_n) = \nu(B\cap F_n)-\nu(A\cap F_n) = \nu((B\setminus A)\cap F_n)$$