There are many questions here about algebras and sigma algebras, but none seem to ask quite what I'm after.
Say we have a whole set $S$, and any collection of subsets $\mathscr{S} \subset 2^S$.
Elementary measure theory that there exists a smallest algebra containing $\mathscr{S}$, and we call it $\alpha(\mathscr{S})$.
We also know that there exists a smallest sigma algebra containing $\mathscr{S}$, and we call it $\sigma(\mathscr{S})$.
That is,
$$
\mathscr{S} \subseteq \alpha(\mathscr{S}) \\
\mathscr{S} \subseteq \sigma(\mathscr{S}) \\
$$
And we also know that the sigma algebra contains the algebra.
$$
\alpha(\mathscr{S}) \subseteq \sigma(\mathscr{S})
$$
But one can also generate a sigma algebra from the algebra, denoted $\sigma(\alpha(\mathscr{S}))$.
My intuition is that the latter sigma algebra contains the former,
$$
\sigma(\mathscr{S}) \subseteq \sigma(\alpha(\mathscr{S}))
$$
Indeed, I suspect that they are equivalent:
$$
\sigma(\mathscr{S}) = \sigma(\alpha(\mathscr{S}))
$$
But I'm having trouble showing or disproving both relationships. Either way, I suspect it's really simple. Is anyone able to help?
Best Answer
Any sigma algebra is an algebra. Hence the smallest algebra containing a collection is necessarily contained in the smallest sigma algebra containing the collection.