Given a set $\Omega$ with its powerset denoted by $\mathscr P(\Omega)$ and a class of its subsets $\mathscr C\subset \mathscr P(\Omega)$:
the generated $\sigma$-algebra $\sigma(\mathscr C)$ is the smallest $\sigma$-algebra which contains $\mathscr C$
Due to the definition of $\sigma$-algebra, $\sigma(\mathscr C)$ always exists (I guess, provided AC) and can be given by the intersection of all $\sigma$-algebras containing $\mathscr C$. However, such implicit construction gives me a hard time when I am trying to understand which properties does $\sigma(\mathscr C)$ have for some particular choices of $\mathscr C$.
Proving the claim in this answer I realized that the following method can be applied. Say, we would like to show that any $A\in \sigma(\mathscr C)$ satisfies a particular property $S$, equivalently it can be stated as $ \sigma(\mathscr C)\subset \mathrm S \tag{1} $ where $\mathrm S\subset \mathscr P(\Omega)$ is the class of all subsets which satisfy $S$. Then we can do the following:
verify that $\mathscr C\subset \mathrm S$ and $\{\emptyset\}\in \mathrm S$ which is a necessary condition;
prove that $\sigma(\mathscr C)\cap \mathrm S$ is a $\sigma$-algebra.
By the minimality of $\sigma(\mathscr C)$ it will imply $(1)$. For example: let $(\Omega,\mathscr F,\mathsf P)$ be a probability space. Define two classes: $ \mathscr F_B = \{A\in \mathscr F:A\subset B\}\text{ for }B\in\mathscr F $ $ \mathscr N = \{A\in \mathscr F:\mathsf P(A) = 0\}. $ For a $\sigma$-algebra $\mathscr F^\prime_B = \sigma(\mathscr F_B\cup\mathscr N)$ I want to show that $ \mathsf P(A\setminus B) \in \{0,1-\mathsf P(B)\}\tag{2} $ for any $A\in \mathscr F^\prime_B$ so, $\mathrm S$ is given by $(2)$. Clearly, $\mathscr F_B\cup \mathscr N\subset \mathrm S$ - and further I show that $\mathscr F^\prime_B\cap\mathrm S$ is closed under taking complements and countable unions.
My issue is the following: I don't know how to apply the same method to the following problem without using results of an example above: let $(\Omega,\mathscr F,\mathsf P) = ([0,1],\mathscr B([0,1]),\lambda)$ where $\mathscr B$ is the Borel $\sigma$-algebra and $\lambda$ is the Lebesgue measure. For the case $B = [0,\frac12]$ I want to show that $\mathscr F^\prime_B$ (defined as above) does not contain sets of the form $[a,1]$ for $a>\frac12$.
Say, I consider $A\in \mathscr F^\prime_B\cap \mathrm S$ so it does not have the form $[a,1]$; hence $A^c\in \mathrm F^\prime_B$ but I don't know how to show that $A^c$ is also not of the from $[a,1]$. With countable unions the problem is similar.
I would appreciate any help with the problem above - also I will be happy to learn about more neat methods of how to verify if the generated $\sigma$-algebra satisfy some property or not.