Complex Lebesgue integral, property Lets say that you for real functions have proved that:
$|\int_{\Omega}fd\mu|\le \int_{\Omega}|f|d\mu$.
How do I then prove that it also holds for complex-valued functions?
I guess this amounts to showing that if $f=u+iv$, then:
$\sqrt{(\int_{\Omega}ud\mu)^2+(\int_{\Omega}vd\mu)^2}\le\int_{\Omega}\sqrt{u^2+v^2}d\mu$.
I guess we see that the property then also holds if the function f is purely imaginary, but what about when it is complex?
 A: Choose $\varphi \in \mathbb{R}$ such that
$$e^{i\varphi}\int_\Omega f\,d\mu \geqslant 0.$$
Then
$$\begin{aligned}
\left\lvert \int_\Omega f\,d\mu\right\rvert
&= e^{i\varphi}\int_\Omega f\,d\mu\\
&= \int_\Omega e^{i\varphi}f\,d\mu\\
&= \int_\Omega \operatorname{Re} (e^{i\varphi}f)\,d\mu\\
&\leqslant \int_\Omega \lvert \operatorname{Re}(e^{i\varphi}f)\rvert\,d\mu\\
&\leqslant \int_\Omega \lvert e^{i\varphi}f\rvert\,d\mu\\
&= \int_\Omega \lvert f\rvert\,d\mu.
\end{aligned}$$
A: HINT: Use Jensen's inequality. 
Works for (semi)norms and vector valued functions. 
Indeed: take a function $f$ with values in a (finite dimensional) normed vector space $(V, ||\cdot||)$. The function $||\cdot||$ is convex so we have the Jensen inequality:
$$ || \frac{1}{\mu(\Omega)} \cdot \int_{\Omega} f\, d\mu || \le \frac{1}{\mu(\Omega)} \cdot \int_{\Omega} ||f||\, d\mu $$
and since $||\cdot ||$ is positively homogenous of degree $1$ we get 
$$ || \int_{\Omega} f\, d\mu || \le   \int_{\Omega} ||f||\, d\mu $$
To handle the case of infinite measure $\mu(\Omega) = \infty$ approximate the integrals with integrals on domains of finite measure.
A: We have:
$$\int_{\Omega}fd\mu=\left|\int_{\Omega}fd\mu\right|(i\sin \theta + \cos \theta)=\left|\int_{\Omega}fd\mu\right|\lambda$$
for $\theta \in (0,2\pi]$ and $\lambda \in \mathbb{C}$, $|\lambda|=1$.Then:
$$\left|\int_{\Omega}fd\mu\right|=\overline{\lambda}\int_{\Omega}fd\mu$$
Next:
$$\Re \left( \left|\int_{\Omega}fd\mu\right|\right)=\Re \left(\overline{\lambda}\int_{\Omega}fd\mu\right)=\int_{\Omega}\Re \overline{\lambda}fd\mu \leq \int_{\Omega}|\Re \overline{\lambda}f|d\mu \leq \int_{\Omega}| \overline{\lambda}f|d\mu=\int_{\Omega}|f|d\mu$$
Because $|\overline{\lambda}|=1$.
