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 $\mu = \nu$.
Answer
Answers can be viewed only if
- The questioner was satisfied and accepted the answer, or
- The answer was disputed, but the judge evaluated it as 100% correct.
1 Attachment
The answer is accepted.
Join Matchmaticians Affiliate Marketing
Program to earn up to 50% commission on every question your affiliated users ask or answer.
- answered
- 292 views
- $7.00
Related Questions
- Black Scholes Calculation
- Question on a pre-measure defined by Folland's real analysis book
- Integral of product measure
- Stochastic Processes Questions
- Finding Probability Density Function of a Standard Brownian motion: Conditioning for two different cases
- Measure Theory and the Hahn Decomposition Theorem
- Probability and Statistics Question help please
- Wiener process probability
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.