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 Radonâ€“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 .