The title is quite misleading, I don't have a better one though. It's clear by definition that $\sigma$-algebra is also an algebra. Here is my question, for those algebras which are not $\sigma$-algebras, say, $\mathcal{A}_0$ on $\Omega$, (where $\mathcal{A}_0$ is an algebra but not a $\sigma$-algebra), what does the following set look like:
$$
\sigma(\mathcal{A}_0)\setminus \mathcal{A}_0?
$$
I'm not quite sure if the answer can be the following:
$$
\sigma(\mathcal{A_0})\setminus \mathcal{A_0}=\{\cup_nA_n:\cup_nA_n\not\in\mathcal{A}_0,A_n\in\mathcal{A}_0\}.
$$
The motivation of this question is from my trying of rediscovering a proof of the extension theorem of probability measure:
Each probability $P$ defined on the algebra $\mathcal{A}_0$ has a unique extension on $\sigma(A_0)$
For the uniqueness part, one may need to prove that two probability measure which agree on $\mathcal{A}_0$ also agree on $\sigma(\mathcal{A}_0)\setminus \mathcal{A}_0$.
Best Answer
There is generally no explicit way to write down how the elements of a generated $\sigma$-algebra look like in the way one can do with the generated algebra. There can be sets that can be constructed that need infinitely many steps to be constructed- and then some more. One can "construct" the generated $\sigma$-algebra in $\omega_1$ steps, with $\omega_1$ being the first uncountable ordinal, by transfinite induction.