Equality of two measures on a generated $\sigma$-algebra.
Let $(\Omega, \mathcal{F})$ be a measurable space and $\mathcal{E} \subset \mathcal{P}(\Omega)$ a generator of $\mathcal{F}$, i.e. $\mathcal{F}$ is the intersection of all $\sigma$-algebras that contain $\mathcal{E}$. Now let $\mu$ and $\nu$ be two measures on $(\Omega, \mathcal{F})$ such that
(i) $\mathcal{E}$ is closed under intersection, i.e.: $A,B \in \mathcal{E} \implies A \cap B \in \mathcal{E}$, and
(ii) $\mu(A) = \nu(A) \quad \forall A \in \mathcal{E}$
Prove that $\mu = \nu$.
Ichbinanonym
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Mathe
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The answer is accepted.
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What do you mean by Pot(Ω)?
I suppose it means Power Set. I believe this advanced question merits a larger bounty.
If you want to know what a fair bounty should be, just think about how much time one may need to spend to answer the question, and how much the time of such individual is worth.
One also needs to assume that \mu(\Omega) = \nu(\Omega)
One also needs to assume that \mu(\Omega) = \nu(\Omega) and that they are finite.