How much larger is the $\sigma$-algebra than the algebra in Caratheodory extension?

Given a 'measure' $\lambda$ on an algebra $\mathcal{A}$ of sets, Caratheodory gives a way to extend this $\lambda$ to a $\sigma$-algebra.

The idea is we define an outer measure (on all subsets) $\lambda^*$ by \begin{equation}\lambda^*(E)=\operatorname{inf} \Sigma\lambda(A_j),\end{equation} where the $\operatorname{inf}$ is taken over all countable coverings of $E$ using sets in $\mathcal{A}$. Then one shows that the collection of subsets satisfying the Caratheodory condition \begin{equation}\lambda^*(E)=\lambda^*(E\cap A)+\lambda^*(E\backslash A)\end{equation} for all $E$ forms a $\sigma$-algebra, say $\mathcal{A}^*$, and $\lambda^*$ restricted to this $\mathcal{A}^*$ is actually a measure.

So my question is, how much larger is this $\mathcal{A}^*$ than $\mathcal{A}$? We know $\mathcal{A}^*$ contains $\sigma(\mathcal{A})$, the $\sigma$-algebra generated by $\mathcal{A}$, and it contains all the null sets. But can it be even larger than this?

In case of the Lebesgue measure, we start with the algebra of all 'boxes' then ended up of the Lebesgue measurable sets, which are in turn union of sets from $\sigma$(boxes) and null sets. So the difference of $\mathcal{A}^*$ and $\sigma(\mathcal{A})$ is exactly the null sets.

But can $\mathcal{A}^*$ be significantly larger than $\sigma(\mathcal{A})$?

Thanks!

Solution 1:

Let $\Omega$ be an uncountable set and $\cal A$ the finite-cofinite field on $\Omega$ (all sets which are either finite or their complement is finite) and $\lambda:{\cal A}\to[0,+\infty]$ counting measure.

Then $\lambda^*$ is counting measure and hence countably additive on all subsets of $\Omega$ which implies that ${\cal A}^*$ is the full power set of $\Omega$.

But $\sigma({\cal A})$ is the countable-cocountable sigma field on $\Omega$ which is significantly smaller than the power set of $\Omega$.

Solution 2:

Systematic Approach:

(Comment for questions!)