Let T be a completely regular space, let ( Ω , F , μ ) (\Omega ,\mathcal {F},\mu ) be complete probability space, and let ρ : L ∞ ( μ ) → L ∞ ( μ ) \rho :{\mathcal {L}^\infty }(\mu ) \to {\mathcal {L}^\infty }(\mu ) be a lifting. If f : Ω → T f:\Omega \to T is a Baire measurable function, must there exist a function f ~ \tilde f with almost all of its values in T, such that ρ ( h ∘ f ) = h ∘ f ~ \rho (h \circ f) = h \circ \tilde f for all bounded continuous functions h on T? If T is strongly measure-compact, then the answer is “yes". If T is not measure-compact, then the answer is “no". This shows that a lifting is not always the best method for the construction of weak densities for vector measures.