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.

Behold the mystery
Behold the mystery

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|<r} |f - y_0| =0   \ \ \ \ \ (1)

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.