Measurability of Banach space valued functions

There is only an indirect proof of the existence of a function {f\colon [0, 1]\to \mathbb R} that is not Lebesgue measurable. But it’s easy to give an explicit example when the codomain of {f} is a Banach space: just let {b(t)} be the sequence of the binary digits of {t}, considered as an element of the sequence space {\ell_\infty}.

Why is {b} not measurable? Recall that a Banach space-valued function {f} is (Bochner) measurable iff there is a sequence of simple functions {\sum v_k \chi_{A_k}} (finite sum, measurable {A_k}, arbitrary vectors {v_k}) that converges to {f} almost everywhere. This property implies that, with an exception of a null set, the range of {f} lies in the separable subspace spanned by all the vectors {v_k} used in the sequence of simple functions. But {b} has the property {\|b(t)-b(s)\|=1} whenever {t\ne s}, so the image of any uncountable set under {b} is nonseparable.

Another way to look at this: on the interval [0, 1) the function {b} is injective and its range has discrete topology, which implies that every subset of [0, 1) is the preimage of some open subset of {\ell_\infty} under {b}.

The binary-digits functions can also be used to illustrate an issue with the duality of Lebesgue-Bochner spaces {L_p(0, 1; X)} where {X} is a Banach space and {1\le p<\infty}. (So, {f} belongs to this space iff it is Bochner measurable and the {L^p} norm of {\|f\|\colon [0, 1]\to [0, \infty)} is finite.) In general we do not have the expected relation {L_p(0, 1; X)^* = L_q(0, 1; X^*)} with {1/p+1/q=1}. The natural isometric embedding of {L_q(0, 1; X^*)} into {L_p(0, 1; X)^*} is still there: any {g\in L_q(0, 1; X^*)} acts on {L_p(0, 1; X)} by {f\mapsto \int \langle f(t), g(t) \rangle\, dt}. But unless {X^*} has the Radon–Nikodym property, these are more bounded linear functionals on {L_p(0, 1; X)}.

To construct such a functional, let {b_n(t)} be the {n}-th binary digit of {t}. Given {f\in L_1(0, 1; \ell_1)}, write it in coordinates as {(f_1, f_2, \dots)} and define {\varphi(f) = \sum_n \int_0^1 f_n b_n}. This is a bounded linear functional, since {|\varphi(f)|\le \sum_n \int_0^1 |f_n| = \|f\|}. But there is no function {g\in L_\infty(0, 1; \ell_\infty)} that represents it, i.e., {\varphi(f) = \int_0^1 \langle f(t), g(t)\rangle \,dt = \sum_n \int_0^1 f_n g_n }. Indeed, if such {g} existed then by considering {f} with only one nonzero coordinate, we find that {g_n} must be {b_n}, using the duality {L_1^* =  L_\infty} in the scalar case. But the function {[0, 1]\to \ell_\infty} with the components {(b_1, b_2, \dots)} is not measurable, as shown above.

This example, which applies to all {1\le p<\infty}, also serves as a reminder that the duality relation {L_p(0, 1; X)^* = L_q(0, 1; X^*)} depends on the dual space {X^*} having the Radon-Nikodym property (RNP), not {X} itself. Indeed, {X=\ell_1} has the RNP; its dual does not.

The importance of {X^*} having the RNP becomes clear once one tries to follow the usual proof of {L_p^*=L_q}. Given {\varphi\in L_p(0,1;X)^*}, we can define an {X^*}-valued measure {\tau} on {[0, 1]} by {\tau(A)(v) = \varphi( v\chi_A)} where {A\subset [0, 1]} is Lebesgue measurable and {v\in X}. This measure has reasonable finiteness and continuity properties coming from {\varphi} being a bounded functional. Still, the existence of a density {g\colon [0, 1]\to X^*} of the measure {\tau} depends on the structure of the Banach space {X^*}.