Real Analysis Folland, Proposition 2.6 Measurable functions Question:

Proposition 2.6 - If $f,g: X\rightarrow \mathbb{C}$ are $M$-measurable, then so are $f + g$ and $fg$.

Attempted proof/brainstorm - Suppose $f,g: X\rightarrow \mathbb{C}$ are $M$-measurable.
We want to show that $f + g\in M$ and $fg\in M$.
Since $f^{-1}$ and $g^{-1}$ preserves unions, intersections, and complements, is there a way of using Proposition 2.3 to show that $f+g$ and $fg$ are $M$-measurable?
I don't really understand Folland's approach. Any suggestions is greatly appreciated.

Background information:

Proposition 2.3 -  If $(X,M)$ is a measurable space and $f:X\rightarrow \mathbb{R}$, the following are equivalent:
a. $f$ is $M$-measurable
b. $f^{-1}((a,\infty))\in M \ \ \forall a\in\mathbb{R}$
c. $f^{-1}([a,\infty))\in M \ \ \forall a\in\mathbb{R}$
d. $f^{-1}((-\infty,a))\in M \ \ \forall a\in\mathbb{R}$
e. $f^{-1}((-\infty,a])\in M \ \ \forall a\in\mathbb{R}$

 A: Response to comment: Yes, that's what I was referencing. Basically, what we want to say is composition of measurable functions is measurable (which is easy), and that, if $\Psi: \mathbb{R}^2\to\mathbb{R}$ is addition, then $\Psi\circ f\times g:X\to\mathbb{R}$ is the sum. You would then have to show that $f\times g$ is measurable, and then you're done. This is doable, but not as direct as one may like.
The clever way to do this is as follows: We want to show that the set of $x\in X$ such that $f(x)+g(x)<a$ for some fixed $a\in\mathbb{R}$ is measurable. This is equivalent to saying $f(x)<a-g(x)$. Now, by density of the rationals, there is an $r\in\mathbb{Q}$ so that $f(x)<r<a-g(x)$. Note we can write this as the intersection of $\{x: f(x)<r\}$ and $\{x: g(x)<a-r\}$.
With this observation, one can show that, $$\{x:f(x)+g(x)<a\}=\bigcup_{r\in\mathbb{Q}}[\{x:f(x)<r\}\cap\{x:g(x)<a-r\}]$$This is a countable union of measurable sets, and is therefore measurable.
A: Let us explain step by step the approach used by Folland.
Let us start with a simpler result

Proposition 2.6 (addition - case Real) - If $f,g: X\rightarrow \mathbb{R}$ are $M$-measurable, then so are $f + g$.

A rather traditional and simple way to prove it is the following.
Proof 1.  Note that, for all $a\in \mathbb{R}$,
\begin{align*}
(f+g)^{-1}((-\infty,a))&=\{x : f(x) + g(x) < a \} = \{x : f(x) < a - g(x)\} =\\
&=\bigcup_{r\in\mathbb{Q}}(\{x : f(x) < r\}\cap \{x : r < a - g(x)\})= \\ 
&=\bigcup_{r\in\mathbb{Q}}(\{x : f(x) < r\}\cap \{x : g(x)  < a - r\})\in M \\
\end{align*}
So $f+g$ is $M$-measurable.
QED
It is immediate that $f-g$ is also $M$-measurable.
This approach is simple for the addition and it is used, for instance in Halmos, Measure Theory. However, this approach is hard to adapt to other ways to combine $f$ and $g$. For instance, to describe $(fg)^{-1}((-\infty,a))$ we will have to consider several sub-cases ($a\geq 0$ or $a<0$, and  pieces where $g$ is negative, positive and zero). To avoid those sub-cases, Halmos uses the smart remark that $$fg=\frac{1}{4}[(f+g)^2-(f-g)^2]$$ and since $f+g$ and $f-g$ are $M$-measurable, we have that $fg$ is $M$-measurable. 
Suppose now the $f$ and $g$ above take values in $\mathbb{C}$. Then the way to use the proof above to prove that $f+g$ is  $M$-measurable is to consider the real and imaginary parts of $f$ and $g$ separately, and use an auxiliary result which states that any function with values in $\mathbb{C}$ is $M$-measurable if and only if its real and imaginary parts are $M$-measurable. In a similar way, considering the real and imaginary parts of $fg$, we can prove that $fg$ is $M$-measurable. 
And this answers one of your questions: Yes, we can prove proposition 2.6 from proposition 2.3, but we need to follow all the steps above.
So, we see that the approach used in proof 1 is simple for addition in Real case, but a little "hard" to adapt to other similar operations. 
That is why some authors (including Folland) prefer a different approach. Here is how it works. Let us show it for the addition and multiplication in the  complex case. 

Proposition 2.6 - If $f,g: X\rightarrow \mathbb{C}$ are $M$-measurable, then so are $f + g$ and $fg$.

Proof 2 - Since $f$ and $g$ are measurable, we know that for any open sets $A, B \subset \mathbb{C}$,   we have $f^-1(A) \in M$ and $g^-1(B) \in M$. 
Let us define $f\otimes g:X\to   \mathbb{C} \times  \mathbb{C}$ such that $f\otimes g \; (x) = (f(x),g(x))$. 
The Borel $\sigma$-algebra in $\mathbb{C} \times  \mathbb{C}$ is generated by the open retangles $A \times B$ where $A$ and $B$ are open subsets of $\mathbb{C}$. 
So to test the $M$-measurability of $f\otimes g$ all we need is to check the preimages of the open rectangles. We have 
$$(f\otimes g)^{-1}(A\times B) = f^{-1}(A) \cap g^{-1}(B)\in M$$
So  $f\otimes g$ is $M$-measurable. 
Now note that if $V$ and $W$ are topological spaces and $H$ is any continuous function from $V$ to $W$, then $H$ is $(B_V,B_W)$-measurable. 
Consider $+ :\mathbb{C} \times  \mathbb{C} \to \mathbb{C}$ (where $+(a,b)=a+b$). It is continuous and so  it is $(B_{\mathbb{C} \times  \mathbb{C}},B_{\mathbb{C}})$-measurable.  So, since $f+g=+(f\otimes g)$, we have that  $f+g$ is $M$-measurable. 
Consider $\times :\mathbb{C} \times  \mathbb{C} \to \mathbb{C}$ (where $\times(a,b)=ab$). It is continuous and so  it is $(B_{\mathbb{C} \times  \mathbb{C}},B_{\mathbb{C}})$-measurable.  So, since $fg=\times(f\otimes g)$, we have that  $fg$ is $M$-measurable.
QED 
This approach is a little bit more complex, more sophisticate, but it promptly applies to all continuous binary operations defined on $\mathbb{R^n}$ or $\mathbb{C^n}$ ($n\geq 1$).
A: Suppose that $\Phi : \mathbb{C}\times \mathbb{C} \to \mathbb{C}$ is continuous. Define $h:X\to \mathbb{C}$ by
$$
h(x) = \Phi(f(x),g(x)).
$$
To prove the measurability of $h$, it suffices to show that the map $F:x\mapsto (f(x),g(x))$ is measurable.
There are many equivalent norms on $\mathbb{C}\times \mathbb{C}$. The convenient one here is
$$
\|(w_1,w_2) - (z_1, z_2)\| = \max \{|w_1-z_1|, |w_2-z_2|\}.
$$
An open ball under this norm has the form $U\times V$, where $U$ and $V$ are open balls in $\mathbb{C}$ (under the usual metric). Since $\mathbb{C}\times \mathbb{C}$ is separable, an open set $W\subset \mathbb{C}\times \mathbb{C}$ is a countable union of open balls $U_n \times V_n$. Hence
$$
F^{-1}(W) = \left(\bigcup_{n=1}^{\infty}f^{-1}(U_n)\right) \cap \left(\bigcup_{n=1}^{\infty}g^{-1}(V_n)\right)
$$
is measurable. That is, $F$ is measurable. Hopefully you know how to apply this to your problem.
