How do you prove $| \int (f)| \leq \int (|f|)$? For an integrable function $f$ on $(a,b)$, how would you prove $| \int (f)| \leq \int (|f|)?$
 A: Let $f$ be a complex measurable function on $(a,b)$. Choose a complex number $\alpha$ such that $\left|\alpha\right|=1$ and $\alpha \int f=\left|\int f\right|$. We now compute that:
$\left|\int f\right|=\alpha \int f = \int \alpha f\leq \int \left|f\right|$. 
Exercise 1: Why does the inequality hold in the above proof? (Hint: observe that the first equality shows that $\int \alpha f$ is a real number.)
Exercise 2: The proof above tells you under what conditions equality holds in the inequality $\left|\int f\right|\leq \int \left|f\right|$. State these conditions. (Hint: equality holds in the inequality given if and only if $\int \alpha f = \int \left|f\right|$.)
I think that it is worth explaining the intuition behind the above proof. The case when $f\geq 0$ is trivial. Let us consider the case $f<0$. In this case, we wish to show that $\left|\int f\right|\leq \int \left|f\right|$. The key point is to use the case $f\geq 0$; we can apply the case $f\geq 0$ to $-f$ and conclude that $\left|\int (-f)\right|\leq \int\left|(-f)\right|=\int (-f)$.  
Exercise 3: Deduce that $\left|\int f\right|\leq \int \left|f\right|$ if $f<0$. 
In the general case, the complex number $\alpha$ above acts in a similar way to the transference of the case $f<0$ to the case $f\geq 0$.
