This is motivated by this question: https://mathoverflow.net/questions/85411/is-there-an-inclusion-of-l-inftyg-into-c-0g and Bill Johnson's comments there.

Let $X$ be a locally compact Hausdorff space and $\mu$ a Radon measure on $X$.  (We can simplify things to make $X$ compact if we like, so "Radon" becomes much the same a "Borel"; but see below).  Suppose $\mu$ has full support.  Let $M^\infty(X,\mu)$ be the space of all bounded $\mu$-measurable functions on $X$; so $L^\infty(X,\mu)$ is the quotient of $M^\infty(X,\mu)$ by the subspace of $\mu$-null functions.
A "strong lifting" $\rho$ is a map $M^\infty(X,\mu) \rightarrow M^\infty(X,\mu)$ which is unital, linear, multiplicative, and such that:

 1. $\rho(f)=f$ $\mu$-a.e.
 2. if $f=g$ $\mu$-a.e. then $\rho(f) = \rho(g)$
 3. if $f$ is continuous, then $\rho(f)=f$.

This means basically that $\rho$ picks out a representative of each equivalence class in $L^\infty(X,\mu)$ (and respects continuous functions) and so allows us to genuinely think of $L^\infty(X,\mu)$ as a space of functions.  Very nice...

However, it seems that there is a hidden technicality.  We might ask that each function $\rho(f)$ actually be Borel.  Apparently this is open, even for $X=[0,1]$ with Lebesgue measure.

So the problem is that $\rho(f)$ might genuinely be only $\mu$-measurable.  In the question I link to above, the hope was that we could use $\rho$ to embed $L^\infty(X,\mu)$ into $C_0(X)^{**}$, but that would require us to be able to integrate $\rho(f)$ against any _bounded_ Radon measure.  So my question is:

> Which measures can we integrate $\rho(f)$ against (for all $f$)?  Assuming we cannot integrate against all measures, is there a good counter-example to illuminate things?

My reference for all of this is the book "Topics in the theory of lifting" by A. and C. Ionescu Tulcea.  This is an old book; I am not an expert.  Has any progress been made on e.g. the Borel measurability question?