Is it possible to work in $L^1$ with a non complete measure?

Question

Basically I think that it is implicit when we traditionally build $L^1(X)$ that the measure is complete. BUT WHAT IF it isn’t complete, is it possible? And what properties do we loss (for sure it won’t be complete but maybe more?)

Answer 1

Let $(X, \mathcal{A}, \mu)$ be a measure space and let $ (X, \overline{\mathcal{A}}, \mu’)$ be its completion. Then the spaces $L^p(X, \mathcal{A}, \mu)$ and $L^p (X, \overline{\mathcal{A}}, \mu’)$ are isomorphic, i.e. for all practical purposes they are the same. Roughly speaking this is because the sigma algebras agree modulo null sets. Hence, whether a measure space is complete or not cannot possibly affect the $L^p$ space, since its measure space completion would give the same $L^p$ space.