Borel $\sigma$-algebra $\mathcal B$ is simply easier to work with (in my opinion) as it behaves well with respect to the topology of our space. Nevertheless, the $\sigma$-algebra $\mathcal M$ of all Lebesgue measurable sets is simply richer. I can conceive that there're probably some interesting stuff going on.
Suppose $\Omega\subset\Bbb R^n$ be open domain. What are some major differences between $L^p(\Omega,\mathcal B,\mu)$ and $L^p(\Omega,\mathcal M,\mu)$?
Does completeness of $\mathcal M$ allow us to prove some interesting theorems that the structure of $\mathcal B$ is not big enough to support? More generally, what do we gain by replacing $\mathcal B$ with its completion in a more general topological space $X$?
In this setting there is no difference at all.
Recall that the elements of $L^p$ are not functions, but equivalence classes of functions, where the equivalence is "equal almost everywhere". It's a standard measure theory fact that every Lebesgue-measurable function is a.e. equal to some Borel-measurable function (this comes from the fact that $\mathcal{M}$ is the completion of $\mathcal{B}$ with respect to $\mu$). So the equivalence classes in $L^p(\Omega, \mathcal{M}, \mu)$ are in one-to-one correspondence with those in $L^p(\Omega, \mathcal{B}, \mu)$, and this correspondence is an isometric isomorphism between the Banach spaces. For all intents and purposes, they are the same space.
The same argument works to compare any measure space with its completion.