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?