In the extended reals, $\mathbb{R}$, let $E$ be a borel subset of $\mathbb{R}$, and let $$ E_1 = E \cup\{\infty\} , E_2 = E \cup\{-\infty\} , E_3 = E \cup\{-\infty, \infty\} . $$ Then $E, E_1, E_2, E_3 $ is a $\sigma$-algebra, with $E$ variying in the borelians.
What I understand of this problem is that I have to show that $\mathcal{F} = \{E, E_1, E_2, E_3 \}$ is a $\sigma$-algebra.
$E \in \mathcal{F} .$ $E$ is any borel set, so it can be $X = \mathbb{R} $ or $\emptyset.$
Take $B \in \mathcal{F}.$ If $B=E,$ as $B$ is in a $ \sigma$-algebra, its complement is also there. So $B^c \in \mathcal{F}.$ (So far, I hope I am getting a correct idea of the proof). But what happens if $B=E_1, $ or $B=E_2, $ or $B=E_3$? I don't see what is the complement of the set whose only element is infinity. $ E_1^c = E^c \cap\{\infty\}^c.$ So I have no idea if this set is in $\mathcal{F}.$ A similar thing happens with $E_2, E_3.$
I would also have to verify that countable unions of elements in $\mathcal{F}$ are in $\mathcal{F}.$ For this I observed various cases: $E \cup...\cup E = E,$ and $E\cup E_1 = E_1,$ $E\cup E_2 = E_2,$ $E\cup E_3 = E_3,$ $E_1\cup E_2 = E_3,$ $E_1\cup E_3 = E_3,$ $E_2\cup E_3 = E_3,$ $E_3\cup E_3 = E_3.$ So for more unions of the sets, I will most likely get $E_3,$ as the result, And $E_3 \in \mathcal{F}.$
For the countable intersection, I am not going to write down all the cases, but the most common intersection of sets will be $E.$ And in other cases, the intersection gives again something in $\mathcal{F}.$
I also beleive I am writing down to much, so I would appreciate an idea of how to simplify the writing, If the Idea of the proof is correct.