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Sometimes it is difficult to explicitely state a $\sigma$-algebra generated by a certain collection of subsets of a set $X$ (sometimes even impossible). I found some answers for the finite case, but what would be the best strategy for example to determine $$\sigma(\{B \subseteq X : A \subseteq B\})$$ for some fixed $A \subseteq X$?

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In this case we have $$\sigma(\{B\subseteq X: A\subseteq B\})=\{C\subseteq X: A\subseteq C\text{ or }C\cap A=\emptyset\}.$$

To see this, for simplicity let $$\Sigma=\{C\subseteq X: A\subseteq C\text{ or }C\cap A=\emptyset\}.$$ First it's easy to prove that $\Sigma$ itself is a $\sigma-$algebra, by just directly verifying three conditions in the definition.

Second, we show that any sigma algebra containing $\{B\subseteq X: A\subseteq B\}$ must contain $\Sigma$. Let $\Sigma'$ be any $\sigma-$algebra containing $\{B\subseteq X: A\subseteq B\}$. Choose any $C\in\Sigma$, then either $A\subseteq C$ or $A\subseteq C^c$, and thus either $C\in \{B\subseteq X: A\subseteq B\}$ or $C^c\in\{B\subseteq X: A\subseteq B\}$. Hence we have $C\in\Sigma'$ and therefore $\Sigma\subseteq \Sigma'$.

Hence the proof is complete.

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    Thanks a lot. Just one question: How did you find the set $\{C\subseteq X: A\subseteq C\text{ or }C\cap A=\emptyset\}$? I mean sometimes I am lacking of intuition to see what it could be. The main problem is finding the set, proving it is another part. – 2017-01-23
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    This really depends on the problem. In general there's no universal strategy to determine the set (as far as I know). Perhaps one way to tackle is by looking at the collection of sets you are given, select some sets, take complements and countable unions and see how the resulting sets look like. At least by doing this you may get some idea of which sets belong to the $\sigma-$algebra generated. If you want to prove certain set is the $\sigma-$algebra generated by a set, then the procudure is pretty much the same as I did. – 2017-01-23