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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?

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

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Any sigma algebra is an algebra. Hence the smallest algebra containing a collection is necessarily contained in the smallest sigma algebra containing the collection.

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  • $\begingroup$ Thank you for the quick response. Sorry if this is really obvious, but would you be able to explain why the second sentence holds? "Hence the smallest algebra containing a collection is necessarily contained in the smallest sigma algebra containing the collection." $\endgroup$ Commented Jan 20, 2020 at 6:06
  • $\begingroup$ @kdbanman It is just the meaning of the word .smallest. The smallest algebra containing a given family is contained in any other algebra containing the given family. $\endgroup$ Commented Jan 20, 2020 at 6:08
  • $\begingroup$ Ah, that helped cement it for me - I did not realize that was the case. I guess we are fortunate that such "smallest" algebras and sigma algebras exist! Thanks again. $\endgroup$ Commented Jan 20, 2020 at 18:33

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