R&C Analysis-Integration of Complex functions

Let $\mu$ be a positive definite measure on arbitarary measurable space $X$. Let $L^{1}(\mu)$ be collection of all complex measurable functions $f$ on $X$ for which \begin{equation} \int_X |f| d\mu < \infty \end{equation} is called space of ``Lesbesgue integral functions''. $f$ measurable implies $|f|$ is measurable, hence above integral is defined. For understanding next definition, recall that a function $u$ can be split into positive $u^+=max\{u,0\}$ and $u^-=-max\{u,0\}$ parts.  
Definition If $f=u+iv$ for real measurable functions $u,v$ on $X$ and if $f \in L^{1}(\mu)$, we define \begin{equation} \int_E f d\mu = \int_E u^+ d\mu - \int_E u^- d\mu + i \int_E v^+ d\mu - i \int_E v^- d\mu \end{equation} for every measurable set $E$.Know that $u^+,u^-,v^+,v^-$ are measurable. Then $\int_E f d\mu$ exists. Futhermore $u^+ \leq |u| \leq |f|$ for all $4$ parts of above integral. Hence, each one of them is finite. Clearly from above definition, $\int_E fd\mu$ is a complex number. Occasionally, it is desirable to define integral of $f$ with range $[-\infty,\infty]$ to be \begin{equation} \int_E f d\mu = \int_E f^+ d\mu - \int_E f^-1 d\mu \end{equation} provided atleast one of the integrals on the right is finite. Thus LHS is a number between $[-\infty,\infty]$. Theorem 1.32 Suppose $f,g \in L^1(\mu)$ and $\alpha,\beta$ are complex numbers, then $\alpha f+\beta g \in L^1(\mu)$ and \begin{equation} \int_X (\alpha f + \beta g) = \alpha \int_X f d\mu +\beta \int_X g d\mu \end{equation} \subsection{Proof} First we need to establish that $\alpha f+\beta g$ is measurable.Then,need to show that the integral is less than infinity - thus establishing that LHS of above belongs to esteemed Legesgue integrable functions set($L^1(\mu)$). If $f,g$ are complex measurable functions, $f+g$ and $fg$ are measurable functions. $\alpha f = f.f.\cdots(\alpha\ times)$ is measurable. Similarly $\beta g$ is measurable. The sum of measurable funcitons $\alpha f + \beta g$ is measurable. WLOG assume $o \leq f \leq g$, then $\int_E f d\mu \leq \int_E g d\mu$. Know, \begin{equation} |\alpha f + \beta g | \leq |\alpha||f| + |\beta||g| \end{equation} This implies \begin{equation} \int_X |\alpha f + \beta g |d\mu \leq \int_X |\alpha||f| + \int_X |\beta||g|=|\alpha|\int_X|f| d\mu + |\beta| \int_X |g| d\mu < \infty \end{equation} Thus $\alpha f + \beta g \in L^1(\mu)$. To prove $(4)$ we need to establish \begin{equation} \int_X (f+g)d\mu = \int_X f d\mu + \int_x g d\mu \end{equation} and \begin{equation} \int_X \alpha f d\mu = \alpha \int_X f d\mu \end{equation} Assume $h=f+g$ \begin{equation} h^+ - h^- = f^+ - f^- + g^+ - g^- \text{ implies } h^++f^-+g^-=f^++g^++h^- \end{equation} From Theorem $1.27$ know that if $f=\sum_{i=1}^\infty f_n(x)$ then $\int_X f \mu = \sum_{i=1}^n \int_X f_n d\mu$. Applying this theorem yields, \begin{equation} \int h^++\int f^-1 + \int g^- = \int f^+ + \int g^+ + \int h^- \end{equation} Since each of these integrals is finite, we can rearrange terms as we like. \begin{equation} \int h^+-\int h^- = \int(h^+-h^-) = \int (f^+- f^-) + \int (g^+- g^-) \end{equation} Leading to $\int_X(f+g) d\mu = \int_X f d\mu + \int_X g d\mu$. To establish equation (8), the following was already proved earlier. \begin{equation} \int_X (\alpha f) d\mu = \alpha \int_X f d\mu \text{ when } \alpha \geq 0 \end{equation} All that's left is to show that equation (8) holds for $\alpha < 0$ and $\alpha=i$. $\alpha=0$ case: Notice \begin{equation} -u^+ = -max\{u,0\} = u^- \end{equation} which means \begin{equation} \int_X (-1) f d\mu = \int_X (-1)(u+iv) d\mu = \int_X(-u-iv)d\mu = (-1)\int_X f d\mu \end{equation} $\alpha=i$ case: \begin{equation} \int (if) = \int i(u+iv) = \int (iu-v) = -\int v + i\int u = i\int (u+iv)=i\int f \end{equation} Proved for all $\alpha$ less than $0$ and for $i$. This shows, \begin{equation} \int_X \alpha f d\mu = \alpha \int_X f d\mu \end{equation}