## 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 ${X^*}$.

## Diagonal

The diagonal ${D=\{(x,x):0\le x\le 1\}}$ of the unit square ${Q=[0,1]\times [0,1]}$ is a mysterious thing. For one thing, its length is an irrational number. But this is not what I’m writing about.

The diagonal has zero area. Lebesgue integrable functions on ${Q}$ form the normed space ${L^1(Q)}$ which, upon closer inspection, consists not of functions but of their equivalence classes. Two functions ${f,g}$ are equivalent if the set ${\{f\ne g\}}$ has zero area. In particular, changing all the value of ${f}$ on the diagonal ${D}$ does not change ${f}$ as an element of ${L^1(Q)}$. The logical conclusion is that given an element ${f\in L^1(Q)}$, we have no way to give a meaning to the integral of ${f}$ over ${D}$.

But wait a moment. If I take two square integrable function ${u,v\in L^2[0,1]}$, then the expression ${\int_0^1 u(x)v(x)\,dx}$ makes perfect sense. On the other hand, it represents the integral of the function ${f(x,y)=u(x)v(y)}$ over the diagonal ${D}$.

Pushing this further, if an element ${f\in L^1(Q)}$ can be represented as ${f(x,y)=\sum_{k=1}^n u_k(x)v_k(y)}$ for some ${u_k,v_k\in L^2[0,1]}$, then ${\int_D f}$ is naturally defined as ${\sum_{k=1}^n \int_0^1 u_kv_k}$. I can even take infinite sums, assuming that everything converges.

This is confusing. If I pick up an element of ${L^1(Q)}$ off the sidewalk, how will I know if it’s safe to integrate it over the diagonal? The existence or nonexistence of the decomposition into sum of products is not obvious.

I guess a satisfactory answer is given by the notion of a Lebesgue point. Given ${f\in L^1(Q)}$ and a point ${p}$ of ${Q}$, consider the following statement:

$\displaystyle \exists y_0\in \mathbb R \text{ such that } \lim_{r\rightarrow 0} \frac{1}{r^2} \int_{|x-p|

The validity of (1) and the value of ${y_0}$ are not affected by the choice of a representative of ${f}$. If (1) holds, ${p}$ is called a Lebesgue point of ${f}$, and we can think of ${y_0}$ as the “true” value of ${f(p)}$ (whether or not our representative agrees with that value). It’s a theorem that almost every point of the domain of ${f}$ is a Lebesgue point. The meaning of “almost every” corresponds to the measure under consideration: on the plane it’s the area, on a line it’s the length.

The product ${f(x,y)=u(x)v(y)}$ has a special feature. Since almost every ${x\in [0,1]}$ is a Lebesgue point of ${u}$, and a.e. ${y\in[0,1]}$ is a Lebesgue point of ${v}$, it follows that almost every point of the diagonal (in the sense of linear measure) is a Lebesgue point of the product ${f}$. (It helps to integrate over small squares instead of disks in (1), which does not change anything.) This makes it possible to define ${\int_D f}$ unambigiously.

The sums of products also have the property that almost every point of ${D}$ is a Lebesgue point. And other elements of ${L^1(Q)}$ may also have this property: they are safe to integrate diagonally, too.