# On $\sigma$-algebra generated by $\mathcal{E}$

Suppose $X$ is a nonempty set. If $\mathcal{E}\subseteq 2^{X}$ (the power set of $X$), the intersection of all $\sigma$-algebras containing $\mathcal{E}$ is called the $\sigma$-algebra generated by $\mathcal{E}$, and is denoted by $\mathcal{M}(\mathcal{E})$. I am trying to prove the following exercise:

If $\mathcal{M}(\mathcal{E})$ is the $\sigma$-algebra generated by $\mathcal{E}$, then $\mathcal{M}(\mathcal{E})$ is the union of the $\sigma$-algebras generated by $\mathcal{F}$ as $\mathcal{F}$ ranges over all countable subsets of $\mathcal{E}$. (Hint: Show that the latter object is a $\sigma$-algebra)

My attempt: Let $\mathcal{N}$ be the union of the $\sigma$-algebras generated by $\mathcal{F}$ as $\mathcal{F}$ ranges over all countable subsets of $\mathcal{E}$. I can prove that $\mathcal{N}$ is a $\sigma$-algebra.

But how do I prove that $\mathcal{E}\subseteq\mathcal{N}$?

If $\mathcal{E}$ is countable, then it is clear. But what happens if $\mathcal{E}$ is uncountable? I appreciate any help.

• See here. – Michael Greinecker Dec 18 '13 at 22:23
• This is probably $\mathcal E\subseteq2^X$, not $\mathcal E\subseteq X$. – Did Dec 18 '13 at 22:25
• @Did: Corrected :-) – Prism Dec 18 '13 at 22:27
• Actually, $\mathcal E\in2^X$ and $\mathcal E\subseteq X$ are equivalent hence $\mathcal E\in2^X$ is wrong as well. – Did Dec 18 '13 at 22:32
• @Did: Ahhh I am blind – Prism Dec 18 '13 at 22:34

Let $E\in\mathcal{E}$.

Then $\mathcal{F}=\left\{ E\right\}$ is a countable subset of $\mathcal{E}$ and $E$ is contained in the $\sigma$-algebra generated by $\mathcal{F}$.

Then also $E\in\mathcal{N}$ (the union).

Proved is now that $\mathcal{E}\subseteq\mathcal{N}$

• So simple! Thank you very much. I was confusing the subsets of $X$ and subsets of $2^{X}$... – Prism Dec 18 '13 at 22:39

Here's a complete proof for the reference. Taken from here. I am making this post community-wiki.

• There is a typo in the beginning of third paragraph. It should say "To see that $\mathcal{M}(\mathcal{E})\subseteq\mathcal{M}$ we first notice that…" I will correct it when I get the chance to latexify the post (right now, it's only a screenshot) – Prism Dec 18 '13 at 23:20