Measurable complex-valued function I have just started a Real Analysis with Measure Theory course. The class uses Bartle's "The Elements of Integration and Lebesgue Measure." We haven't covered much so my tools are pretty limited. I am finding the following problem a bit hard, because we have only seen measurable functions $f:X \to \mathbb{R},$ and the problem deals with a measurable function $f:X \to \mathbb{C}.$ I think I should think of $\mathbb{C}$ as $R^2$ (but still we haven't covered anything in $R^d$ for $d>1.$)

Let $f$ be a complex-valued function defined on a measurable space ($X$, $\Sigma$). Show that $f$ is $\Sigma-$measurable if and only if
$\{x \in X:a<Re(f)(x)<b,c<Im(f)(x)<d\}$
belongs to $\Sigma$ for all real numbers $a,b,c,d$. More generally, $f$ is $\Sigma-$measurable if and only if $f^{-1}(G)\in \Sigma$ for every open set $G$ in the complex plane.

This is the definition of measurable function that I have: $f$ is $\Sigma$-measurable if $\forall a \in \mathbb{R}$, $f^{-1}(a,\infty) \in \Sigma$. (The book also proves that we can substitute $\{(a,\infty): a \in \mathbb{R}\}$ with any other generating family of $B(\mathbb{R}))$.
In terms of a $\Sigma$-measurable complex-valued function, I have read and re-read the $2$ chapters we have covered so far and there is only one paragraph on $\Sigma$-measurable complex-valued functions, which reads: If $f$ is a complex-valued function defined on $X$ then there exist two uniquely determined real-valued functions $f_r$ and $f_i$ such that $f=f_r+if_i, f_r(x)=Ref(x)$ and $f_i(x)=Imf(x).$ We define the complex-valued function to be $\Sigma$-measurable if and only if its real and imaginary parts $f_r$ and $f_i$ are measurable.
This is what I have so far:
Let $R=I_1 \times I_2$ where $I_1$ and $I_2$ are open intervals in $\mathbb{R}$. Suppose $f$ is $\Sigma$-measurable, then $f_r$ and $f_i$ are $\Sigma$-measurable $\implies f_r^{-1}(I_1) \in \Sigma$ and $f_i^{-1}(I_2) \in \Sigma$.
I know it's not much, but I don't know how to continue because I don't know how to deal with $\mathbb{R}^2.$ I understand this is too much to ask but is there anyone who would be willing to write me a detailed proof (without skipping steps or writing "then it follows...") with explanations and using very basic tools. I would like to understand the proof well so I can get rid of my confusion.
 A: Suppose $f$ is complex measurable. Then $\operatorname{re} f$ and $\operatorname{im} f$ are measurable in the ordinary sense and hence $(\operatorname{re} f)^{-1} ((a,b)) \cap (\operatorname{im} f)^{-1} ((c,d))$ is measurable.
Now suppose $(\operatorname{re} f)^{-1} ((a,b)) \cap (\operatorname{im} f)^{-1} ((c,d))$ is measurable for all $a,b,c,d$. Since $\mathbb{R} = \cup_n (-n,n)$ we see that 
$\cup_n ((\operatorname{re} f)^{-1} ((a,b)) \cap (\operatorname{im} f)^{-1} ((-n,n))) = (\operatorname{re} f)^{-1} ((a,b))$ is measurable and so $\operatorname{re} f$ is measurable. Similarly, $\operatorname{im} f$ is measurable and so $f$ is measurable.
Note:
From the point of view of measurability, $\mathbb{C}$ is treated as if it was $\mathbb{R}^2$ and you can deal with this in at least two equivalent ways.
A real valued function is considered (Borel) measurable iff $\{x| f(x) \le \alpha \}$ is measurable for all $\alpha$ ($<,>,\le,\ge$ are all equivalent in this regard).
(i) Bartle calls a function $f: X \to \mathbb{C}$ measurable iff the functions $\operatorname{re} f$ and $\operatorname{im} f$ are real measurable.
(ii) You can call $f: X \to \mathbb{C}$ measurable iff $\{x | f(x) \in (-\infty, \alpha] \times (-\infty,\beta] \}$ is measurable for all $\alpha, \beta$.
It is not too hard to show that these are the same.
