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.