Let $\mathcal{U}$ be a random class of subsets from $\Omega $.
Through
$$\sigma (\mathcal U) = \bigcap_{\mathcal U \subseteq \mathcal{F}\text{ and } \mathcal{F} \text{ is a } \sigma\text{-algebra on } \Omega} \mathcal{F} $$
a $\sigma$ algebra is given, i.e. the minimal $\sigma$ algebra, which contains $\mathcal U$.
How can one prove that $\sigma(\mathcal U)$ is really a $\sigma$-algebra?
This is what I did. Let $U$ and $V$ denote two subsets of a nonempty set $\Omega$. Let $\mathcal F$ denote the smallest sigma-algebra containing $U$ and $\mathcal G$ the smallest sigma-algebra containing $V$.
- If $\mathcal F\subset \mathcal V$ and $\mathcal F\ne \mathcal V$. Then, either $U=\varnothing$ or $U=\Omega$.
- If $\mathcal F=\mathcal U$. Then, either $U=V$ or $U=\Omega\setminus V$.
But I don't think that's the answer to what is being asked for.
I don't know how I can check the three properties (https://en.wikipedia.org/wiki/%CE%A3-algebra#Definition) on what is given above.
Best Answer
In general, intersections tend to behave nicely when every member of the intersection is closed under an operation. For example, $\emptyset$ is contained in every such $\mathcal{F}$ so it is contained in their intersection, $\sigma(U)$. Using the same argument you can guarantee that the other properties hold. For the closure under complementation, let $S \in \sigma(U)$. Then, by definition, $S$ belongs to every member $\mathcal{F}$ of the intersection. Since $\mathcal{F}$ is a $\sigma$-algebra we have $\Omega \setminus S \in \mathcal{F}$ so $$\Omega \setminus S \in \bigcap_{\;\quad U \subseteq \mathcal{F} \\ \mathcal{F} \text{ is a }\sigma\text{-algebra}} \mathcal{F} = \sigma(U).$$
Can you finish the proof from here?