## 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 ${X^*}$.

## Almost norming functionals, Part 2

Let $E$ be a real Banach space with the dual $E^*$. Fix $\delta\in (0,1)$ and call a linear functional $e^*\in E^*$ almost norming for $e$ if $|e|=|e^*|=1$ and $e^*(e)\ge \delta$. In Part 1 I showed that in any Banach space there exists a continuous selection of almost norming functionals. Here I will prove that there is no uniformly continuous selection in $\ell_1$.

Claim. Let $S$ be the unit sphere in $\ell_1^n$, the $n$-dimensional $\ell_1$-space.  Suppose that $f\colon S\to \ell_{\infty}^n$ is a map such that $f(e)$ is almost norming $e$ in the above sense. Then the modulus of continuity $\omega_f$ satisfies $\omega_f(2/n)\ge 2\delta$.

(If an uniformly continuous selection was available in $\ell_1$, it would yield selections in $\ell_1^n$ with a modulus of continuity independent of $n$.)

Proof. Write $f=(f_1,\dots,f_n)$. For any $\epsilon\in \{-1,1\}^n$ we have $n^{-1}\epsilon \in S$, hence

$\sum\limits_{i=1}^n \epsilon_i f_i(n^{-1}\epsilon)\ge n\delta$ for all $\epsilon\in \{-1,1\}^n$. Sum over all $\epsilon$ and change the order of summation:

$\sum\limits_{i=1}^n \sum\limits_{\epsilon}\epsilon_i f_i(n^{-1}\epsilon)\ge n2^n\delta$

There exists $i\in\{1,2,\dots,n\}$ such that

$\sum\limits_{\epsilon}\epsilon_i f_i(n^{-1}\epsilon) \ge 2^n \delta$

Fix this $i$ from now on. Define $\tilde \epsilon$ to be the same $\pm$ vector as $\epsilon$, but with the $i$th component flipped. Rewrite the previous sum as

$\sum\limits_{\epsilon} -\epsilon_i f_i(n^{-1}\tilde \epsilon)\ge 2^n\delta$

$\sum\limits_{\epsilon}\epsilon_i [f_i(n^{-1}\epsilon)-f_i(n^{-1}\tilde \epsilon)]\ge 2^{n+1}\delta$
Since $\|n^{-1}\epsilon-n^{-1}\tilde \epsilon\|=2/n$, it follows that $2^n \omega_f(2/n) \ge 2^{n+1}\delta$, as claimed.
Let $E$ be a real Banach space with the dual $E^*$. By the Hahn-Banach theorem, for every unit vector $e\in E$ there exists a functional $e^*\in E^*$ of unit norm such that $e^*(e)=1$. One says that $e^*$ is a norming functional for $e$. In general, one cannot choose $e^*$ so that it depends continuously on $e$. For example, the 2-dimensional space with $\ell_1$ norm does not allow such a continuous selection.
Fix $\delta\in (0,1)$ and call a linear functional $e^*$ almost norming for $e$ if $|e|=|e^*|=1$ and $e^*(e)\ge \delta$. In any Banach space there exists a continuous selection of almost norming functionals.