Real Analysis, Folland Proposition 2.21 Integration of Complex Functions 
Proposition 2.21 - The set of integrable real-valued functions on $X$ is a real vector space, and the integral is a linear functional on it.

Attempted proof - Note that we can derive the axioms of a vector space by proving that for $a,b\in \mathbb{R}$ and $f,g\in L^1$ we have $|af + bg|\leq |a||f| + |b||g|$. So, \begin{align*}
|af + bg| = |a(f^+ + f^-) + b(g^+ + g^-)| &\leq |a(f^+ - f^-)| + |b(g^+ + g^-)|\\
&= |a||f^+ + f^-| + |b||g^+ + g^-|\\ &= |a||f| + |b||g|
\end{align*}
Now to complete the first assertion we need to show that $\int a f = a\int f$. Let $a\in\mathbb{R}$ and $f\in L^1$ then $$\int |af| = \int |a(f^+ + f^-)| = \int |a||f^+ + f^-| = |a|\int |f^+ + f^-| = |a|\int |f| $$
Now, to show additivity, suppose $f$ and $g$ are integrable and $h = f+g$. Then $h^+ - h^- = (f^+ - f^-) + (g^+ - g^-)$, so, $h^+ + f^- + g^- = h^- + f^+ + g^+$. Then applying Theorem 2.15 $$\int h^+ + \int f^- + \int g^- = \int h^- + \int f^+ + \int g^+$$ Since $h = h^+ - h^-$ rearranging we get $$\int h = \int h^+ - \int h^- = \int f^+ - \int f^- + \int g^ + - \int g^- = \int f + \int g$$
I am not sure if this is correct, any suggestions is greatly appreciated.
 A: I have made the required adjustments /corrections in your proof.  

Proposition 2.21 - The set of integrable real-valued functions on $X$ is a real vector space, and the integral is a linear functional on it.

Proof - To prove that the set of integrable real-valued functions on $X$ is a real vector space, it is enough to prove that for all $a,b\in \mathbb{R}$ and all $f,g\in L^1$ , $af+bg \in L^1$. But this is an immediate consequence of $af+bg$ being measurable and $|af + bg|\leq |a||f| + |b||g|$.
(proof that  $|af + bg|\leq |a||f| + |b||g|$: 
$ \:\:\: |af + bg|\leq |af| + |bg|= |a||f| + |b||g|$)
Now let us prove the integral is linear. We begin proving  $\int a f = a\int f$. Let $a\in\mathbb{R}$ and $f\in L^1$ then, if $a> 0$, $$\int af = \int (af)^+ -\int(af)^-=  \int af^+ - \int af^-=a \left(\int f^+ - \int f^-\right) =a\int f $$
if $a < 0$, $$\int af = \int (af)^+ -\int(af)^-=  \int |a|f^- - \int |a|f^+=|a |\left(\int f^- - \int f^+\right) = \\=-|a |\left(\int f^+ - \int f^-\right) =-|a|\int f=a \int f $$
if $a=0$,  $$\int af = \int 0 = 0 = a \int f$$
Now, to show additivity, suppose $f$ and $g$ are integrable and $h = f+g$. Then $h^+ - h^- = (f^+ - f^-) + (g^+ - g^-)$, so, $h^+ + f^- + g^- = h^- + f^+ + g^+$. Then applying Theorem 2.15 $$\int h^+ + \int f^- + \int g^- = \int h^- + \int f^+ + \int g^+$$ Since $h = h^+ - h^-$ rearranging we get $$\int h = \int h^+ - \int h^- = \int f^+ - \int f^- + \int g^ + - \int g^- = \int f + \int g$$
