proving $|\int_X{}fd\mu|\leq\int_X{}|f|d\mu$? Let $(X,\Sigma_x,\mu)$ be a measurable space ($X$-the set, $\Sigma_x$ - is a $\sigma$ algebra over X, and $\mu$ is a positive measure).
Let $f\in L^1(\mu)$ (meaning $\int_X|f|d\mu<\infty$), $f:X\to \mathbb{C}$.
How do I prove $|\int_X{}fd\mu|\leq\int_X{}|f|d\mu$?

my thoughts:
Denote $z=\int_Xfd\mu$, let $a\in\mathbb{C}$ be such that $az=|z|$ and then:
$|\int_X{}fd\mu|=|z|=az=a\int_X{}fd\mu$
but...then what?...
any help? :)
 A: In spirit, this is just the triangle inequality, and that's how we'll prove it. First assume $f$ is real valued. Remember that
$\int f d\mu = \sup_g \int g d\mu$
where the supremum is taken over all simple functions $g \leq f$, i.e. all $g$ of the form 
$g(x) = \sum_j c_j 1_{A_j}$
for some finitely many real numbers $c_j$ and measurable sets $A_j$. The inequality holds for any such $g$, since
$\Big| \int g d\mu \Big| = \Big| \sum_j c_j \mu(A_j) \Big| \leq \sum_j |c_j|\mu(A_j) = \int |g| d\mu,$
by the triangle inequality. Thus
$\Big| \int f d\mu \Big| \leq \sup_g \Big| \int g d\mu \Big| \leq \sup_g \int |g| d\mu = \int |f| d\mu.$
Note that the assumption $f \in L^1(\mu)$ is unnecessary, since in the case that $f$ is not integrable, the right hand side is infinite, so the inequality holds trivially. 
If $f$ is complex valued, then take $\alpha \in \mathbb{C}$ so that $\alpha \int f \in \mathbb{R}$, and write
$\Big| \int f d\mu \Big| = \int Re(\alpha f) d\mu \leq \int |Re(\alpha f)| d\mu \leq \int |\alpha f| d\mu = \int |f| d\mu$. 
Edit: There are a couple of not-completely-trivial facts I've used. It would be good to check these to make sure you understand what's going on!
A: Hope this helps:
Proposition: If $f\in L^{1}$, then $\left|\int f\right| \leq \int |f|$.
Proof - If $f$ is real then $$\left|\int f\right| = \left|\int f^{+} - \int f^{-}\right| \leq \int f^{+} + \int f^{-} = \int |f|$$
If $f$ is complex-valued and $\int f\neq 0$, then $\int f = \left|\int f\right|e^{-i\theta}$. So $$\left|\int f\right| = e^{-i\theta}\int f = \int e^{-i\theta}f$$ In particular, $\int e^{-i\theta}f$ is real, so we have 
\begin{align*}
\left|\int f\right| = Re\int e^{-i\theta}f = \int Re(e^{-i\theta} f) &\leq \int |Re(e^{-i\theta}f)|\\
&\leq \int |e^{-i\theta} f|\\
&= \int |f|
\end{align*}
Some of the notation above is from Real Analysis by Folland, if you have any questions please let me know.
