There is only an indirect proof of the existence of a function that is not Lebesgue measurable. But it’s easy to give an explicit example when the codomain of
is a Banach space: just let
be the sequence of the binary digits of
, considered as an element of the sequence space
.
Why is not measurable? Recall that a Banach space-valued function
is (Bochner) measurable iff there is a sequence of simple functions
(finite sum, measurable
, arbitrary vectors
) that converges to
almost everywhere. This property implies that, with an exception of a null set, the range of
lies in the separable subspace spanned by all the vectors
used in the sequence of simple functions. But
has the property
whenever
, so the image of any uncountable set under
is nonseparable.
Another way to look at this: on the interval [0, 1) the function is injective and its range has discrete topology, which implies that every subset of [0, 1) is the preimage of some open subset of
under
.
The binary-digits functions can also be used to illustrate an issue with the duality of Lebesgue-Bochner spaces where
is a Banach space and
. (So,
belongs to this space iff it is Bochner measurable and the
norm of
is finite.) In general we do not have the expected relation
with
. The natural isometric embedding of
into
is still there: any
acts on
by
. But unless
has the Radon–Nikodym property, these are more bounded linear functionals on
.
To construct such a functional, let be the
-th binary digit of
. Given
, write it in coordinates as
and define
. This is a bounded linear functional, since
. But there is no function
that represents it, i.e.,
. Indeed, if such
existed then by considering
with only one nonzero coordinate, we find that
must be
, using the duality
in the scalar case. But the function
with the components
is not measurable, as shown above.
This example, which applies to all , also serves as a reminder that the duality relation
depends on the dual space
having the Radon-Nikodym property (RNP), not
itself. Indeed,
has the RNP; its dual does not.
The importance of having the RNP becomes clear once one tries to follow the usual proof of
. Given
, we can define an
-valued measure
on
by
where
is Lebesgue measurable and
. This measure has reasonable finiteness and continuity properties coming from
being a bounded functional. Still, the existence of a density
of the measure
depends on the structure of the Banach space
.