It is clear that $\left|\displaystyle\int fd\mu\right|\leq\displaystyle\int|f|d\mu$ in any general setting of measure space $(X,\Sigma,\mu)$. But it is necessary that, for any $\delta>0$, there exists some $A_{\delta}\in\Sigma$ such that $\left|\displaystyle\int_{A_{\delta}}fd\mu\right|>(1-\delta)\displaystyle\int_{X}|f|d\mu$?
At least I think this is true in Euclidean Lebesgue integrals, as I have seen some authors take it for granted.
One may have to choose $A_{\delta}$ to be "switching" the negative part of $f$ to the positive one to do the job, but I am struggling to make it.
Any idea?
Edit:
So how does the author manage to choose such a $B_{n}$ to make the integral greater than the $1/4$ of the absolute integrand one?
Not true. Let $\Omega =\{1,2\}$ with the power set and the counting measure. Let $f(1)=1$ and $f(2)=-1$. Then $\int |f|=2$. For any subset $A$ of $\Omega$ we have $\int_A f$ is $0$ or $1$. So the inequality fails whenever $\delta <\frac 1 2$.