I am having difficulty proving the following,

Let $\displaystyle f: X \rightarrow R \ { } f \in L^{1}(\mu) \ { } \alpha \in R$

Prove that $\displaystyle \int_{X}{ \alpha f}d \mu \ { } = \alpha \int_{X}{f} d\mu$

Background: $\displaystyle L^{1}(\mu)$ is the set of complex measurable functions f such that $\displaystyle \int_{X}{|f|} d\mu < \infty $
I will also be using $\displaystyle f^{+}=Max\{f,0\} \ { } f^{-}=Max\{-f,0\}=-Min\{f,0\}$ and $\displaystyle f=f^{+}-f^{-}$ and a lemma that if f is a positive measurable function then
$\displaystyle f,g \geq 0 \ { } \Rightarrow \int (f+g)d\mu =\int {f}d\mu + \int g d\mu $

My Proof thus far: (I have separated the proof into 2 cases)

Case 1: $\displaystyle \alpha \geq 0$

$\displaystyle \int_{X}\alpha f d \mu =\ { } \int_{X}\alpha(f^{+}-f^{-})d \mu \ { } \int \alpha(f^{+})- \alpha(f^{-}) d\mu = \int (\alpha f)^{+} -(\alpha f^{-}) d\mu$

and this is where I am stuck both $\displaystyle \alpha f :X \rightarrow [0,\infty]$ so I can apply the lemma stated earlier but I can not split the integral unless both functions are positive but $\displaystyle -(\alpha f^{-}$ is negative.

Any suggestions on how i can get around this obstacle would be greatly appreciated !