Complement of a set and inverse image. I'm currently taking a real analysis class and we are working on measurable functions (the notes can be found here under "Measurable Functions"; the exercise is the first one on the last page). 

$$\text{Let }f:E \to \mathbb{R}\text{ and }A\subseteq \mathbb{R}.\\\text{
Show that }f^{-1}(A^{C}) = E - f^{-1}(A).$$

I can easily visualize that this is indeed true (I drew a graph), but I'm unsure how to approach the proof. This chapter is dealing with measurable functions, however we do not have in the hypothesis that $f$ is measurable, and I'm not sure this is even needed in proving the claim. 
My first strategy was assuming in two cases that $A$ is closed and open (therefore $A^{C}$ is open and closed respectively) and then going from there using neighborhoods, but this route seems like I'm making the problem harder than it is. I'm also unsure if I could prove the claim this way. 
I'm also curious if I can prove this equality by basic set inclusion, i.e.: an element from set $A$ is in set $B$ and vice versa. 
Any assistance in approach would be greatly appreciated. 
EDIT: Also, I believe this could be a simple property of preimages and such, but still I'm unsure how to approach. 
 A: We need the

Definition of $f^{-1}(C)$:
$$f^{-1}(C)=\{x\in E|\ f(x)\in C\}.$$

Then we just compute both sides and compare.
On one hand,
$$f^{-1}(A^c)=\{x\in E| f(x)\in A^c\}=\{x\in E| f(x)\in \mathbb{R}\setminus A\}$$
On the other hand,
$$E\setminus f^{-1}(A)=E\setminus \{x\in E| f(x)\in A\}=\{x\in E| f(x)\in \mathbb{R}\setminus A\}$$
A: Here's another way to think about it that might help.
Let $g$ be an indicator function defined as follows.
$$ \begin{align} g(x) = 1 & \;\; \text{if and only if} \;\; f(x) \in A \\ g(x) = 0 & \;\; \text{otherwise} \end{align} $$
$f^{-1}(A^C)$ is $g^{-1}(0)$ and $E - f^{-1}(A)$ is $E - g^{-1}(1)$ . Since the image of $g$ is $\{0, 1\}$ everything that isn't sent to $1$ is sent to $0$. Therefore $g^{-1}(0)$ is equal to $E - g^{-1}(1)$. Therefore, as desired
$$ f^{-1}(A^C) = E - f^{-1}(A)$$
Another way of looking at it is, since $f^{-1}(A^C)$ and $f^{-1}(A)$ are disjoint and $f$ is a total function on $E$ (presumably), the following is true because, forall $x$ in $E$, $f$ either sends $x$ to $A$ or it doesn't.
$$ f^{-1}(A^c) \sqcup f^{-1}(A) = E $$
$\sqcup$ denotes the disjoint union.
