- #1
rukawakaede
- 59
- 0
Homework Statement
Let [tex](\Omega,\mathcal{F},P)[/tex] be a probability space. Let [tex]\mathcal{N}\subseteq\mathcal{P}(\Omega)[/tex] where [tex]\mathcal{N}:=\{N\subseteq\Omega:\exists M\in\mathcal{F},P(M)=0[/tex] and [tex]N\subseteq M\}[/tex].
(a) Want to show that [tex]\mathcal{F}'=\{F\cup N:F\in\mathcal{F}, N\in\mathcal{N}\}[/tex] defines a [tex]\sigma[/tex]-field. **only closed under complement part.
(b) For [tex]F,F'\in\mathcal{F}[/tex] and [tex]N,N'\in\mathcal{N}[/tex] with [tex]F\cup N=F'\cup N'[/tex] we have [tex]P(F)=P(F')[/tex].
* I need some directions on how to approaching these.
Homework Equations
All information given at (1.)
The Attempt at a Solution
(a) this part I only need assistance on the complement closure part. My approach is first show the empty set is in [tex]\mathcal{N}[/tex] then show that [tex]F'^c=(F\cup N)^c=F^c\cap N^c\in\mathcal{F}'[/tex]. Initially I thought I would be just fine to use Continuity from below to show that [tex]N^c=\Omega[/tex] but this is completely false, since [tex]N[/tex] can be non empty and it is in [tex]\mathcal{N}[/tex] not in [tex]\mathcal{F}[/tex]. Continuity lemma does not apply.
(b) for this part I made a fault direct attempt. I intended to show [tex]F\cup N=F'\cup N'[/tex] implies [tex]P(F\cup N)=P(F'\cup N')[/tex] but this is totally wrong too since [tex]P[/tex] is not define for [tex]N\in\mathcal{N}[/tex].
Thanks in advanced.