I think they're trying to say:
$$ \|f\|_p=\left(\int_\Omega \vert f\vert^p dv\right)^{1/p} $$ One way to justify their shorthand is that integration is like "measuring" the function, so $v\vert f\vert^p$ could be thought of as shorthand for $\int_\Omega\vert f\vert^pdv$.