# On the smallest sigma algebra of a set

Let $X$ be any set. Say $\mathcal{E} \in 2^X$. Then there exists a unique smallest $\sigma-$algebra containing $\mathcal{E}$

### Attempt:

Put $$\mathcal{C} = \{ \mathcal{F} : \mathcal{F} \; \text{is a sigma algebra} \; \; and \; \; \mathcal{E} \subset \mathcal{F} \}$$

$\mathcal{C}$ is non-empty since the sigma algebra $2^X$ lives in there trivially. Next, write

$$\mathcal{S} = \bigcap_{C \in \mathcal{C} } C$$

$\mathcal{S}$ is a sigma algebra since intersection of sigma algebras is a sigma algebra. Can someone help me to show why this set is the smallest unique sigma algebra? I am stuck. Thanks.

• These are the indications you were given. Now you are supposed to check that the class S satisfies the axioms of a sigma-algebra. Did you try at least some of them? – Did Nov 25 '14 at 9:18
• I did check that. I got stuck in the part to show minimality and uniqueness. That is why I asked the question. Also, I don't understand the downvote. – ILoveMath Nov 25 '14 at 9:32
• We cannot say what you checked and what you did not, where you had trouble and where you had not, since you do not say. Note that, now that you accepted an answer recalling only well-known definitions, we still cannot say. Thus, the benefit for the site is assuredly near zero and the benefit for you remains quite mysterious. – Did Nov 25 '14 at 10:15

If $\mathcal{F}$ is a $\sigma$-algebra with $\mathcal{E}\subseteq\mathcal{F}$ then $\mathcal{F}\in\mathcal{C}$ so that by definition $\mathcal{S}:=\bigcap\mathcal{C}\subseteq\mathcal{F}$.
Characteristic for the smallest $\sigma$-algebra containing $\mathcal E$ is that is a subcollection of every $\sigma$-algebra containing $\mathcal E$.
If $\mathcal{F}_{1}$ and $\mathcal{F}_{2}$ are both the smallest $\sigma$-algebra containing $\mathcal{E}$ then $\mathcal{F}_{1}\subseteq\mathcal{F}_{2}$ because $\mathcal{F}_{2}$ is a $\sigma$-algebra containing $\mathcal{E}$ and $\mathcal{F}_{1}$ will - as smallest $\sigma$-algebra containing $\mathcal E$ - be a subcollection of $\mathcal{F}_{2}$ . Likewise $\mathcal{F}_{2}\subseteq\mathcal{F}_{1}$ so that $\mathcal{F}_{1}=\mathcal{F}_{2}$. This proves uniqueness.