Proof $\sigma (U) = \bigcap_{U \subset \mathcal{F}, \mathcal{F} - \sigma - \text{Algebra of } \Omega} \mathcal{F} $ is sigma-algebra 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.

Let it be that $\mathcal F_i$ is a $\sigma$-algebra on $\Omega$ for every $i\in I$.
Then it can be proved straightforwardly that $\bigcap_{i\in I}\mathcal F_i$ is a $\sigma$-algebra on $\Omega$.
By straighforwardly I mean that you only have to check that the collection is closed under complements and under countable unions (and is not empty).
Not really difficult and it shows immediately that the $\sigma(\mathcal U)$ defined in your question is a $\sigma$-algebra on $\Omega$.
