Sigma-algebra defined by intersection

Question

Let $F$ be a system of sets thus that $F=\lbrace A \rbrace$. Where A is subset of M that is not specified. Then $F=\lbrace \emptyset, A, A^c, M\rbrace$. Suppose further that we have the definition of $\sigma(F)$ which is based on the intersections of all $\sigma$-algebras containing $F$. How can I give an explanation that this delivers me $\sigma(F)=\lbrace \emptyset, A, A^c, M\rbrace$?

Answer 1

I will change the notation a bit to ease the reading a bit.

Let $\Omega$ be your sample space and $A\subset\Omega$. You can then show that $\sigma(A)$ is the smallest $\sigma$-algebra (up to inclusion) which contains $A$.

Then by definition we have $$A \in \sigma(A), A^c\in\sigma(A), \Omega\in \sigma(A), \emptyset\in\sigma(A)$$

The latter $2$ follow because a $\sigma(A)$ is closed under complements and contains $\Omega$, while the first $2$ because $A\in\sigma(A)$ and again we have that it's closed under complements.

Now back to your problem - I feel like you are missing a bit of brackets in what you want to prove that the sigma algebra contains but that might be just me since your $F$'s seem different to me. Otherwise my proof is directly applicable.

Answer 2

If $\sigma(F)$ is defined as in your question then it is not difficult to prove that $\sigma(F)$ is a $\sigma$-algebra.

Characteristic for $\sigma$-algebra $\sigma(F)$ is then: $$\text{ if }\mathcal A\subseteq\wp(M)\text{ is a }\sigma\text{-algebra with }F\subseteq\mathcal A\text{ then }\sigma(F)\subseteq\mathcal A\tag1$$

This follows directly from the definition.

So you could say that $\sigma(F)$ is the smallest (or minimal) $\sigma$-algebra that contains $F$.

Based on the fact that $\sigma(F)$ is a $\sigma$-algebra with $F=\{A\}\subseteq\sigma(F)$ it can be deduced that: $$\{\varnothing,A,A^{\complement},M\}\subseteq\sigma(F)\tag2$$

It is not difficult to show that $\{\varnothing,A,A^{\complement},M\}$ is a $\sigma$-algebra with $F\subseteq\{\varnothing,A,A^{\complement},M\}$.

Now from $(1)$ it follows that:$$\sigma(F)\subseteq\{\varnothing,A,A^{\complement},M\}\tag3$$ Combining $(2)$ and $(3)$ we find:$$\sigma(F)=\{\varnothing,A,A^{\complement},M\}$$