extension of the property of a set to its generated $\sigma$-algebra Theorem: Suppose that $(\Omega,\mathcal{F})$ is a measurable space and that $c \subset \mathcal{F}$ is such that

*

*$c$ contains $\Omega$.

*$c$ is closed under intersection.(i.e.$A \cap B \in c$, if $A,B \in c$)

If $\mu$ and $v$ are a pair of finite measures on $(\Omega,\mathcal{F})$ and $\mu(A)=v(A)$ for each $A \in c$, then $\mu(A)=v(A)$ for any $A \in \sigma(c)$.
Prove this Theorem.
I have tried to construct a $\mathcal{F_1}=\{A \subset \mathbb{R}|\mu(A)=v(A)\}$. If $\mathcal{F_1}$ is a $\sigma$-algebra, then the theorem can be easily proved.
Obviously,
$\mathcal{F_1}$ is closed under complementary and $\phi,\Omega \in \mathcal{F_1}$.
Therefore, we only need to show that $\mathcal{F_1}$ is countably additive. But I don't know how to prove it or how to use the second property of $c$.
 A: Partial answer: It is enough to show that $\mathcal{F}_1$ is closed under finite unions.
To prove that the set is closed under countable unions observe that if $A_1,A_2,...\in \mathcal{F}_1$ are disjoint then $$\mu(\bigcup A_n ) = \sum_n \mu(A_n) = \sum_n \nu(A_n) = \nu(\bigcup A_n)$$
Now let  $A_1,A_2,...,$ countably many sets in your family (not necessarily disjoint).
Then consider the following sets
Let $A_1' = A_1$ and by induction let $A_i' = A_i \backslash \bigcup_{j=1}^{i-1} A_i$
by finite additivity you get that all $A_i'$ are in your set, also they are disjoint and $\bigcup_{i=1}^\infty A_i = \bigcup_{i=1}^\infty A_i'$. This completes the proof assuming that you know that the family is closed under finite unions.
For finite unions I don't know what to do. But observe that using a proof by induction it is enough to show that if $A$ and $B$ are in $\mathcal{F}$ then $A\cup B$ is in $\mathcal{F}$. Since $\mu(A\cup B) = \mu(A)+\mu(B)-\mu(A\cap B)$ it is enough to show that $\mathcal{F}$ is closed under finite intersections.
