I am self-studying measure theory using Measure Theory by Donald Cohn. I got confused by the following two remarks made in this book:
Example 2.6.5$\quad$ Let $(X,\mathscr{A})$ be a measurable space, and let $f$ be an $\mathbb{R}^d$-valued function on $X$. Let $f_1,\dots,f_d$ be the components of $f$, i.e., the real-valued functions on $X$ that satisfy $f(x) = (f_1(x),f_2(x),\dots,f_d(x))$ at each $x$ in $X$. Then Proposition 2.6.2 and part (2) of Proposition 1.1.5 imply that $f$ is measurable with respect to $\mathscr{A}$ and $\mathscr{B}(\mathbb{R}^d)$ if and only if $f_1,\dots,f_d$ are $\mathscr{A}$-measurable. It follows from this remark and Proposition 2.1.5 and 2.1.7 that the class of measurable functions from $(X,\mathscr{A})$ to $(\mathbb{R}^d,\mathscr{B}(\mathbb{R}^d))$ is closed under the formation of sums, scalar multiples, and limits.
Example 2.6.6$\quad$ Now consider the space $\mathbb{R}^2$, and identify it with the set $\mathbb{C}$ of complex numbers. The remarks just above imply that a complex-valued function on $(X,\mathscr{A})$ is measurable with respect to $\mathscr{A}$ and $\mathscr{B}(\mathbb{C})$, that is, with respect to $\mathscr{A}$ and $\mathscr{B}(\mathbb{R}^2)$, if and only if its real and imaginary parts are $\mathscr{A}$-measurable, and that the collection of measurable functions from $(X,\mathscr{A})$ to $(\mathbb{C},\mathscr{B}(\mathbb{C}))$ is closed under the formation of sums and limits and under multiplication by real constants. Similar arguments show that the product of two measurable complex-valued functions on $X$ is measurable; in particular, the product of a complex number and a complex-valued measurable function is measurable.
The definition of measurability with respect to two arbitrary measurable spaces $(X,\mathscr{A})$ and $(Y,\mathscr{B})$ as well as Proposition 2.6.2, 1.1.5, 2.1.5, and 2.1.7 are presented at the bottom of this post.
My questions are:
In example 2.6.5, why wouldn't the class of measurable function from $(X,\mathscr{A})$ to $(\mathbb{R},\mathscr{B}(\mathbb{R}^d))$ be closed under the formation of multiplications and quotients? I think Proposition 2.1.7 also implies these, right?
As for Example 2.6.6, I am baffeled by the claim "the product of two measurable complex-valued functions on $X$ is measurable". First of all, what is the product of two complex valued functions? (I have absolutely no background in complex variables other than the definition of addition and multiplication of complex numbers: if $x=(a,b)$ and $y=(c,d)$, then $x+y = (a+c,b+d)$ and $xy = (ac-bd, ad+bc)$.) If the multiplication of two complex valued functions are defined in the same way as the multiplication of two complex numbers, then why are we allowed to identify $\mathbb{C}$ with $\mathbb{R}^2$? The product of two complex number is NOT the inner product of two elements in $\mathbb{R}^2$. So, I am confused by how would one prove the product of two measurable complex-valued functions on $X$ is measurable by forming a similar argument?
Definition of measurability with respect to two measurable spaces $(X,\mathscr{A})$ and $(Y,\mathscr{B})$$\quad$ Let $(X,\mathscr{A})$ and $(Y,\mathscr{B})$ be measurable spaces. A function $f:X\to Y$ \textit{measurable with respect to $\mathscr{A}$ and $\mathscr{B}$} if for each $B$ in $\mathscr{B}$ the set $f^{-1}(B)$ belongs to $\mathscr{A}$. Instead of saying that $f$ is measurable with respect $\mathscr{A}$ and $\mathscr{B}$, we will sometimes say that $f$ is a \textit{measurable function} from $(X,\mathscr{A})$ to $(Y,\mathscr{B})$ or simply that $f:(X,\mathscr{A})\to(Y,\mathscr{B})$ is \textit{measurable}. Likewise, if $A$ belongs to $\mathscr{A}$, a function $f:A\to Y$ is \textit{measurable} if $f^{-1}(B)\in\mathscr{A}$ holds whenever $B$ belongs to $\mathscr{B}$.
Proposition 2.6.2$\quad$Let $(X,\mathscr{A})$ and $(Y,\mathscr{B})$ be measurable spaces, and let $\mathscr{B}_0$ be a collection of subsets of $Y$ such that $\sigma(\mathscr{B}_0) = \mathscr{B}$. Then a function $f:X\to Y$ is measurable with respect to $\mathscr{A}$ and $\mathscr{B}$ if and only if $f^{-1}(B)\in\mathscr{A}$ holds for each $B$ in $\mathscr{B}_0$.
Proposition 1.1.5$\quad$The $\sigma$-algebra $\mathscr{B}(\mathbb{R}^d)$ of Borel subsets of $\mathbb{R}^d$ is generated by each of the following collections of sets:
(1) the collection of all closed subsets of $\mathbb{R}^d$;
(2) the collection of all closed half-spaces in $\mathbb{R}^d$ that have the form $\{(x_1,\dots,x_d):x_i \leq b\}$ for some index $i$ and some $b$ in $\mathbb{R}$;
(3) the collection of all rectangles in $\mathbb{R}^d$ that have the form\begin{align*}\{(x_1,\dots,x_d):a_i < x_i \leq b_i\ \textit{for}\ i = 1,\dots,d \}. \end{align*}
Proposition 2.1.5$\quad$Let $(X,\mathscr{A})$ be a measurable space, let $A$ be a subset of $X$ that belongs to $\mathscr{A}$, and let $\{f_n\}$ be a sequence of $[-\infty,+\infty]$-valued measurable functions on $A$. Then
(1) the functions $\sup_nf_n$ and $\inf_nf_n$ are measurable,
(2) the functions $\limsup_{n\to\infty}f_n$ and $\liminf_{n\to\infty}f_n$ are measurable, and
(3) the function $\lim_{n\to\infty}f_n$ (whose domain is $\{x\in A:\limsup_{n\to\infty}f_n(x)=\liminf_{n\to\infty}f_n(x)\}$) is measurable.
Proposition 2.1.7$\quad$Let $(X,\mathscr{A})$ be a measurable space, let $A$ be a subset of $X$ that belongs to $\mathscr{A}$, let $f$ and $g$ be measurable real-valued functions on $A$, and let $\alpha$ be a real number. Then $\alpha f$, $f+g$, $f-g$, $fg$, and $\frac{f}{g}$ (where the domain of $\frac{f}{g}$ is $\{x\in A:g(x)\neq0\}$) are measurable.
Thank you very much!