In my measure theory lectures notes, I am given a proof for this inequality:
If $f$ is a measurable function such that $f: X \rightarrow \mathbb{C}$, then $$ |\int f d\mu | \leq \int |f| d \mu$$
This is the proof:
$\exists \lambda \in \mathbb{C}$ such that $|\lambda|=1$ and $$|\int f d \mu | = \lambda \int f d\mu = \int \lambda f d \mu \text{ (1)}$$ Let $g = \mathcal{Re}(\lambda f)$, then $|g| \leq |\lambda f| \leq |f| $, so: $$\underset{\in \mathbb{R}}{|\int f d \mu|} = \underset{\in \mathbb{R}}{\int gd \mu} + i \underset{= 0}{\int \mathcal{Im}(\lambda f) d \mu} \text{ (2)}$$ $$= |\int f d \mu| \leq \int |g|d \mu \leq \int |f| d \mu$$
Does the $\lambda $ in (1) is equal to $\overline{\int f d\mu}$? Because that's the only way I can see that $|\int f d \mu | = \lambda \int f d\mu$ as $|\int f d \mu | \in \mathbb{R}$. But I don't really see why then would $|\lambda| = 1$.
And secondly, why in (2) de we have $\int \mathcal{Im}(\lambda f) d \mu = 0$?
Ad 1: No. Whenever $z_1,z_2$ are complex numbers of same absolute value, there exists $\lambda\in\Bbb C$ such that $z_1=\lambda z_2$. Namely, we can take $\lambda=\frac{z_2}{z_1}$ provided $z_1\ne 0$. Here, $z_2$ is the integral and $z-1$ its absolute value (which happens to be real, but still is of same absolute value).
Ad 2: The left hand side is real, the first summand on the right is real, hence the second summand must be real. At the same time it is purely imaginary. Therefore it must be zero.