Extensions of universal measures Let $(\Omega,\mathcal F)$ be a measurable space, and let $\mathcal P$ be the set of all probability measures no this space. Let $\mathcal F^p$ denote a completion of $\mathcal F$ w.r.t. $p\in P$ and define $\mathcal F' = \bigcap_{p\in P}\mathcal F^p$.
Given any $p\in P$, by extension we obtain a unique measure $e(p)$ on $\mathcal F'$. Conversely, given a measure $p'$ on $\mathcal F'$ we can define its restriction to $\mathcal F$ as some $r(p') \in P$. Is that true that $r\circ e$ and $e\circ r$  are identity maps? If this is not true in general, does it hold for $\Omega$ being a standard Borel space, $\mathcal F$ being its Borel $\sigma$-algebra, hence $\mathcal F'$ begin the universal $\sigma$-algebra?

Yes, this follows by unwrapping the definitions. Let's recall that (the restriction of) the completion $e(p)$ is defined as follows: if $F'\in\mathcal F'\subset\mathcal F^p$, then we can write $F'=F\cup N$ with $N\subset N'$ and $F,N'\in\mathcal F$ and $p(N')=0$. In this situation, $e(p)(F')=p(F)$. (That this is well defined, is part of the standard theory.)
Now let's consider $e(r(p'))(F')$, with $F'\in\mathcal F'$. As above, write $F'=F\cup N$ with $N\subset N'$ and $F,N'\in\mathcal F$ and $r(p')(N')=p'(N')=0$. Then, by definition of the completion of a measure, $e(r(p'))(F')=r(p')(F)=p'(F)$. However, since $F'\setminus F\subset N'$ and $p'(N')=0$, we also have that $p'(F)=p'(F')$, so indeed $e(r(p'))=p'$.
Finally, $r(e(p))=p$ is trivial.
