Understanding $\sigma$-algebra with a collection of subsets

Question

So I am taking a graduate course in real analysis, and we just covered sigma algebras. I thought I had it down, but now I'm a bit confused about part of it.

Suppose that $A \subset \Omega$. Then we have that $\sigma(\{A\}) = \{\Omega,\emptyset,A,A^c \}$. This is fine and makes sense to me, but the issue is when we take a collection of subsets $\mathcal{C}$. How then would we define $\sigma(\mathcal{C})$? I know that each subset in this collection will have that same property I have mentioned, but what is a good way to think about what this is doing?

Answer 1

Let $A\subset \mathcal{P}(X)$, then $\sigma(A)$ is the smallest sigma algebra that contains $A$ (in the sense that if $A\subset \sigma$, then $\sigma(A)\subset \sigma$). This definition has a couple of problems. How do we know such a $\sigma(A)$ even exists? To make sense of this, let me prove this fact:

Arbitrary intersection of sigma-algebras is still a sigma algebra!

Indeed, $\cap_{i\in I}\sigma_i$ is such that $\emptyset$ and $X$ are in each $\sigma_i$ so they must be in the intersection. Likewise, if $X_1$ is in every $\sigma_i$, $X_1^C$ is also in each $\sigma_i$ (and hence in the intercection of these sigma algebras). Same thing if$X_i \in \cap \sigma_i$ for every natural $i$, because we must have $\cup_{i\in I} X_i$ in every $\sigma_i$ and therefore in the intersection aswell.

There is a sigma algebra that contains $A$.

Nothing mysterious here, we have $A\subset \mathcal{P}(X)$ and $\mathcal{P}(X)$ is a $\sigma$ algebra...

Okay, now we are ready to really understand why our definition is well posed:

$\sigma(A)$ exists and is given by $\cap_{A\subset \sigma_i}\sigma_i$, the intersection of all sigma algebras that contain $A$.

Indeed, we are actually taking the intersection of sets that exist, because $\mathcal{P}(X)=\sigma_{i_o}$ for example. If $A\subset \sigma$ then we must have $\sigma$ in the intersection and $\cap_{A\subset \sigma_i}\sigma_i\subset \sigma $ . This means $U=\cap_{A\subset \sigma_i}\sigma_i$ is the smallest sigma algebra that contains $A$.

But why are such sets unique. Why couldn't there be another smallest set? Suppouse, $\sigma(A)$ is another set with this property. But then $U\subset \sigma (A)$, because $U$ is one of the possible smallest set that contains $A$ and $\sigma(A)\subset U$ because $\sigma(A)$ is also one of the smallest set that contains $A$. But then $\sigma(A)=U$.