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
116
Answer
Answers can only be viewed under the following conditions:
- The questioner was satisfied with and accepted the answer, or
- The answer was evaluated as being 100% correct by the judge.
1 Attachment
Mathe
3.5K
The answer is accepted.
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- answered
- 815 views
- $7.00
Related Questions
- Measure Theory and the Hahn Decomposition Theorem
- How do we describe an intuitive arithmetic mean that gives the following? (I can't type more than 200 letters)
- Prove that $\frac{d \lambda}{d \mu} = \frac{d \lambda}{d \nu} \frac{d \nu}{d \mu}$ for $\sigma$-finite measures $\mu,\nu, \lambda$.
- A question in probability theory
- Measure Theory (A counterexample to interchanging limits and integration)
- Wiener process probability
- Black Scholes Calculation
- Card riffle shuffling
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.