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This is motivated by this question: Is there an inclusion of $L_\infty(G)$ into $C_0(G)^{**}$? 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?

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A modern reference for all things lifting is chaperter 34 in the epos of Fremlin: According to the following, the existence of a Borel lifting is independent of ZFC:… – Michael Greinecker Jan 13 '12 at 16:43
On existence of Borel lifting (independent of ZFC) see the first paragraph here – Gerald Edgar Jan 13 '12 at 16:46
@Gerald, @Michael: Thanks! – Matthew Daws Jan 13 '12 at 18:06

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