Complete, standard Borel spaces? (Complete Borel isomorphic probability spaces)

Question

A common assumptions in probability theory is that the probability space is complete, i.e. if $\left(\Omega, \mathcal{A}, \mathbb{P}\right)$ is a probability space and $A \in \mathcal{A}$ satisfies $\mathbb{P}\left(A \right) = 0$ and $B \subset A$, then $B \in \mathcal{A}$. Another common assumption is that the $\left(\Omega, \mathcal{A} \right)$ is Borel isomorphic or standard (I hope I'm using the right terms here); there exists a bi-measurable map $\varphi: \left(\Omega, \mathcal{A}\right) \to \left( \mathbb{R}, \mathcal{B} \left(\mathbb{R}\right) \right)$. My question is if there is any contradiction (or if one can run into trouble) if one assumes that both holds, i.e. that one has a complete, Borel isomorphic probability space? Does this relate to the notion of Lebesque-Rokhlin-spaces somehow?