Proving D is a Semi-Ring and Identifying \sigma (D) in [0,1]

[johnson123]

Homework Statement

Let D be the collection of all finite subsets ( including the empty set) of [0,1].
Prove that D is a semi-ring. What is \sigma(D) ? Define on D: \mu (A)=#A . Prove that \mu is a premeasure and identify \mu_{e} and
\Sigma_{mu_{e}} . Is ([0,1],\sigma (D), \mu_{e}) complete?
Prove that ([0,1],\sigma (D), \mu_{e}) \neq
([0,1],\Sigma_{mu_{e}},\mu_{e}).

Homework Equations

\mu_{e} is the outer measure,
\Sigma_{mu_{e}} is the collection of all \mu_{e} measurable sets.

\sigma (D) is the sigma algebra generated by D

The Attempt at a Solution

showing that D is a semi ring is clear.
but \sigma (D) is a little unclear, since it must be closed under complementation, so if A \in D, then A is a finite set, but A^{c}
may not be a finite set.
showing that \mu is a pre-measure is clear.
any comments for the rest is appreciated.

 

[Mystic998]

Um, the complement of a finite set in [0,1] will definitely not be finite. But why is that a problem? Doesn't it just make the sigma algebra generated by D the collection of all sets with finite complement in [0,1] and their complements?

Edit: Oh, sorry, it has to be closed under countable unions, so I guess it's not that simple.

 

[morphism]

Why stick to finiteness? sigma algebras work well with countability. The sigma algebra generated by D certainly contains all countable sets and sets whose complement is countable (i.e. cocountable sets); can it contain anything else?