Indicator function, $\sigma$-algebra, unpacking statements. This is relevant to my question here.

Suppose $\mathcal{F}$ is a collection of real-valued functions on $X$ such that the constant functions are in $\mathcal{F}$ and $f + g$, $fg$ and $cf$ are in $\mathcal{F}$ whenever $f$, $g \in \mathcal{F}$ and $c \in \mathbb{R}$. Suppose $f \in \mathcal{F}$ whenever $f_n \to f$ and each $f_n \in \mathcal{F}$. Define the function$$\chi_A(x) = \begin{cases} 1 & \text{if }x \in A \\ 0 & \text{if }x \notin A.\end{cases}$$Prove that $\mathcal{A} = \{A \subset X : \chi_A \in \mathcal{F}\}$ is a $\sigma$-algebra.

I don't know even to begin with this monster. For one, I'm having a hard time visualizing what's going on. What does $\mathcal{F}$ even look like? What does $\mathcal{A}$ even look like? Could anybody just simply explain to me what all the relevant terms here look like/unpack the relevant terms here, and from there, how to get started?
 A: You have some set $X$ and a collection $\mathcal{F}$ of functions on $X$. You're told that $\mathcal{F}$ is closed under certain operations on functions, like addition and scalar multiplication. Now you're going to show that the closure properties of $\mathcal{F}$ guarantee that $\mathcal{A}$ has some nice closure properties as well.
In particular, you will show that $\mathcal{A}$ is a $\sigma$-algebra. Recall what this means: $X \in \mathcal{A}$ and $\mathcal{A}$ is closed under complementation and countable unions (and hence countable intersections as well, by DeMorgan's laws), i.e. if $A \in \mathcal{A}$, then $A^{c} \in \mathcal{A}$ and if $\{ A_{n} \}_{n \in \mathbb{N}} \in \mathcal{A}$, then $\cup_{n}A_{n} \in \mathcal{A}$. 
I'll get you started with checking some of the easy stuff. 
How to show $X \in \mathcal{A}$? Well, by the definition of $\mathcal{A}$, $X \in \mathcal{A}$ iff $\chi_{X} \in \mathcal{F}$. But of course $\chi_{X} \in \mathcal{F}$ because $\chi_{X} = 1$ is constant on $X$. So $X \in \mathcal{A}$.
Now for complementation. Suppose $A \in \mathcal{A}$. We want to show that $\chi_{A^{c}} \in \mathcal{F}$, for then, by the definition of $\mathcal{A}$, we'll have $A^{c} \in \mathcal{A}$, thus proving that $\mathcal{A}$ is closed under complementation. The trick is to write $\chi_{A^{c}}$ using $\chi_{A}$ (which is in $\mathcal{F}$ by supposition) and the operations under which $\mathcal{F}$ is closed. Consider the function $-(\chi_{A} - 1)$. Can you finish the proof?
