Let $(X,A,\mu)$ be a measure space and $f,f',g,g'$ functions in $\mathcal{L}^1(X,A,\mu)$ such that $f=f'$ $\mu$-almost everywhere and $g=g'$ $\mu$-almost everywhere.
How to prove that $$\int|f-g|d\mu=\int|f'-g'| d\mu?$$
What I know:
I know that there is a theorem that states that $\int fd\mu=0$ iff $f=0$ $\mu$-almost everywhere. So $f-f'-(g-g')=0$ $\mu$-almost everywhere which means that $\int |f-f'-(g-g')|d\mu=0$.
How do I get $\int|f-g|d\mu=\int|f'-g'| d\mu?$
Equality almost everywhere is a property stable by addition and multiplication. With your hypothesis, you know that $f-g=f'-g'$ a.e. . It is also stable under absolute value, thus $|f-g|=|f'-g'|$ a.e., that is $|f-g|-|f'-g'|=0$ a.e. . You apply your theorem, apply additivity of the integrale.
If you don't know stability under absolute value, I think itcomes basically from a decomposition $f=f^+-f^-$.