Real Analysis, Folland The Dominated Convergence Theorem Background Information:

Proposition 2.16 - If $f\in L^+$, then $\int f = 0$ iff $f = 0$ a.e.

Question:

2.24 The Dominated Convergence Theorem - Let $\{f_n\}$ be a sequence in $L^1$ such that (a) $f_n\rightarrow f$, and (b) there exists a nonnegative $g\in L^1$ such that $|f_n|\leq g$ a.e. for all $n$. Then $f\in L^1$ and $\int f = \lim_{n\rightarrow \infty}\int f_n$.

Attempted proof - Suppose $\{f_n\}$ be a sequence in $L^1$ such that (a) $f_n\rightarrow f$, and (b) there exists a nonnegative $g\in L^1$ such that $|f_n|\leq g$ a.e. for all $n$. It is obvious that $f\in L^1$. Now since $f\in L^1$ and $L^1$ is redefined to be the set of equivalence classes of a.e.-defined integrable functions on $X$. Then let $E$ be the set where $f$ is not an integrable function on $X$ thus $\mu(E) = 0$ by definition of a.e. Then $f\in L^1$ for $X\setminus E$. By the Monotone Convergence Theorem $$\lim_{n\rightarrow \infty}\int_{X\setminus E} f_n = \int_{X\setminus E}f$$ Observe that $\int_{E}f = \int_{X}f\chi_E = 0$ by Proposition 2.16. Similarly, $\int_E f_n = 0$. Thus we have \begin{align*}
\int_{X}f &= \int_{X\setminus E}f + \int_{E}f\\
&= \lim_{n\rightarrow \infty}\int_{X\setminus E}f_n + \lim_{n\rightarrow \infty}\int_E f_n\\
&= \lim_{n\rightarrow \infty}\left(\int_{X\setminus E}f_n + \int_E f_n\right)\\
&= \lim_{n\rightarrow \infty}\int f_n
\end{align*}
I am not sure this is right since this result may only apply to $L^+$ (the space of all measurable functions from $X$ to $[0,\infty]$). Any suggestions is greatly appreciated.
 A: @Wolfy, It is true: since we can deduce Fatou's lemma from the Monotone Convergence Theorem,  and the Dominate Convergence Theorem is a consequence of Fatou's lemma, we can prove the Dominate Convergence Theorem from the Monotone Convergence Theorem. However, such path would make us to prove again (inside the proof of the Dominate Convergence Theorem) results that we already have in Fatou's lemma.

2.24 The Dominated Convergence Theorem - Let $\{f_n\}$ be a sequence in $L^1$ such that (a) $f_n\rightarrow f$ a.e., and (b) there exists a nonnegative $g\in L^1$ such that $|f_n|\leq g$ a.e. for all $n$. Then $f\in L^1$ and $\int f = \lim_{n\rightarrow \infty}\int f_n$.

Proof - Let $\{f_n\}$ be a sequence in $L^1$ such that  $f_n\rightarrow f$ a.e. So there is  $E\in \mathcal{M}$, such that $\mu(E)=0$ and for all $x \in E^c$, $f_n(x) \to f(x)$. So, we have  $f_n\chi_{E^c} \to  f\chi_{E^c}$. So  $f\chi_{E^c}$ is measurable and  $f\chi_{E^c}=f$ a.e., by equivalence relation $= a.e.$  we can say $f$ is measurable (in precise terms: the equivalence class of $f$ has a measurable representant).    
We had to be careful in the paragraph above, because $\mu$ is not assume to be complete, so proposition 2.21b does not hold. 
Let us now look at the Real case first. 
Suppose $f_n$'s and $f$ are real-valued. Since $|f_n|\leq g $ a.e.,  we have that $f\in L^1$ and we also have $ 0\leq f_n+g \leq 2 g$. Using Fatou's Lemma (including the Reverse Fatou's Lemma for $\limsup$), we have 
$$\int (f +g) = \int \liminf_n (f_n + g)\leq   \liminf_n\int (f_n + g) \leq \limsup_n\int (f_n + g) \leq \int \limsup_n (f_n + g)= \int (f +g) $$
So we have $\int (f +g) =\lim_n\int (f_n + g)$. Since $\int g <\infty$, we have 
$$\int f = \int (f +g) - \int g =\lim_n\int (f_n + g) -\int g = \lim_n\int f_n$$
Let us now look at the Complex case. 
Suppose $\{f_n\}$ be a sequence in $L^1$ such that $f_n\rightarrow f$ a.e., and there exists a nonnegative $g\in L^1$ such that $|f_n|\leq g$ a.e. for all $n$.
Then  $\textrm{Re}(f_n)\rightarrow \textrm{Re}(f)$ a.e. and, for all $n$, 
$$ |\textrm{Re}(f_n)| \leq |f_n|\leq g$$
So by the result for the Real case, we have  $\textrm{Re}(f) \in L^1$ and $\int \textrm{Re}(f) = \lim_n\int\textrm{Re}(f_n)$. 
In a similar way, we prove that  $\textrm{Im}(f) \in L^1$ and $\int \textrm{Im}(f) = \lim_n\int\textrm{Im}(f_n)$. 
It follows that  $f \in L^1$ and $\int f = \lim_n\int f_n$. 
