# Is the smallest algebra containing a collection always within the smallest sigma algebra containing that collection?

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?

• Yes, by necessity, since any $\sigma$ algebra containing $\alpha(S)$ contains $S$. Commented Jan 20, 2020 at 5:49