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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.

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    $\begingroup$ The problem is not really about measure theory, but about set-theory. $\endgroup$ – OR. Apr 1 '15 at 15:49
  • $\begingroup$ This definitely narrows it a bit--thanks! $\endgroup$ – Tadpole Apr 1 '15 at 15:51
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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, $$E\setminus f^{-1}(A)=E\setminus \{x\in E| f(x)\in A\}=\{x\in E| f(x)\in \mathbb{R}\setminus A\}$$

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  • $\begingroup$ Thanks for the answer, I think I was getting too hung up on thinking about measure, and what we have previously learned (about open and closed sets). $\endgroup$ – Tadpole Apr 1 '15 at 16:10

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