# What's the difference between algebra and $\sigma$-algebra?

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 $\sigma(\mathcal{A}_0)\setminus \mathcal{A}_0$.

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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.
For example, let $\mathcal{A}_0$ be the algebra generated by the open sets of $\mathbb{R}$. Then $\sigma(\mathcal{A}_0)$ is the $\sigma$-algebra of Borel sets. It is much larger than the set of countable unions of elements of $\mathcal{A}_0$. That set of countable unions is indeed not even an algebra. For some details, you may want to read about the Borel Hierarchy.