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 $\mu = \nu$.
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What do you mean by Pot(Ω)?
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I suppose it means Power Set. I believe this advanced question merits a larger bounty.
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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.
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One also needs to assume that \mu(\Omega) = \nu(\Omega)
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One also needs to assume that \mu(\Omega) = \nu(\Omega) and that they are finite.
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