Radon

Nikodým & Banach

Krein

Bourgain

Talagrand

$F(A) = \int_A f \, d\mu$

$C = \overline{\mathrm{co}}\big(\operatorname{ext}(C)\big)$

$S(A, f, \alpha) = \{ x \in A : f(x) > M(A, f) - \alpha \}$

$\mathbb{E}(f_{n+1} \mid \Sigma_n) = f_n$

$x \notin \overline{\mathrm{co}}\big(A \setminus B(x, \varepsilon)\big)$

$|x| = x \vee (-x)$

$\mathrm{RNP} \iff \mathrm{KMP}$

$x = \sum_{i=1}^n \lambda_i x_i, \quad \lambda_i \geq 0, \quad \sum_{i=1}^n \lambda_i = 1$

In a Banach lattice, the Radon–⁠Nikodým property and the Krein–⁠Milman property are equivalent

The following is a friendly exposition of the paper

Dans Un Espace de Banach Reticule Solide, La Propriété de Radon-Nikodým et Celle de Krein-Milman Sont Équivalentes by Jean Bourgain and Michel Talagrand.

If you have any comments or corrections,
please contact the author at:

David Muñoz-Lahoz
<david [dot] munnozl [at] uam [dot] es>

1. Introduction

Two classical properties of Banach spaces are the Radon–Nikodým property (RNP), which characterizes when the Radon–Nikodým theorem holds for vector-valued measures, and the Krein–Milman property (KMP), which characterizes when the Krein–Milman theorem holds for closed bounded convex sets. It turns out that not only does the RNP admit a geometric characterization, but also the RNP implies the KMP. It is one of the major open problems in the geometry of Banach spaces whether the converse holds.

In [6], J. Bourgain and M. Talagrand show that the converse does hold for Banach lattices, and provide two proofs of it: one using existing geometric machinery, and a more direct one using the order structure. This result is remarkable, since it derives a purely geometric theorem using the order structure of the space. Moreover, its proof uses the geometry of Banach spaces, the theory of order continuous Banach lattices, and martingales, which are also closely related to the RNP.

Our goal is to provide a gentle exposition of [6], starting with a smooth introduction to the RNP in Section 2, the KMP in Section 3, and Banach lattices in Section 4, all before jumping to the main contents of the paper in Section 5.

2. The Radon–Nikodým property

Let us start by recalling the classical Radon–Nikodým theorem.

Theorem 2.1 (Radon–Nikodým). Let $(\Omega, \Sigma, \mu)$ be a $\sigma$-finite measure space and let $\nu \colon \Sigma \to \R$ be a finite signed measure that is absolutely continuous with respect to $\mu$, that is, $\nu(A) = 0$ whenever $A \in \Sigma$ satisfies $\mu(A) = 0$. Then there exists $f \in L_1(\mu)$ such that \[\nu(A) = \int_A f \, d\mu \quad \text{for every } A \in \Sigma.\] Moreover, $f$ is unique up to $\mu$-almost everywhere equality.

Informally, a Banach space $X$ satisfies the Radon–Nikodým property if the analogue of the Radon–Nikodým theorem holds for measures with values in $X$. Thus, a precise formulation of this property requires the notion of $X$-valued measures and of integrals of $X$-valued functions. Throughout, $X$ denotes a Banach space and $(\Omega, \Sigma, \mu)$ a finite measure space.

Definition 2.2. A vector measure is a map $F \colon \Sigma \to X$ that is countably additive, that is, for every sequence $(A_n)$ of pairwise disjoint sets in $\Sigma$, $\sum_n F(A_n)$ converges and it equals $F(\bigcup_n A_n)$.

The variation of $F$ is the map $|F| \colon \Sigma \to [0, \infty]$ given by \[|F|(A) = \sup \Big\{ \sum_{i=1}^n \|F(A_i)\| : A_1, \ldots, A_n \in \Sigma \text{ is a partition of } A \Big\},\] and $F$ is of bounded variation if $|F|(\Omega) < \infty$. We say that $F$ is $\mu$-continuous if $F(A) = 0$ whenever $A \in \Sigma$ satisfies $\mu(A) = 0$.

There are several natural ways to integrate $X$-valued functions; in our case, we are interested in the Bochner integral. A function $s \colon \Omega \to X$ is simple if it is of the form $s = \sum_{i=1}^n x_i \chi_{A_i}$ with $n \in \N$, $x_i \in X$ and $A_i \in \Sigma$; its integral over $A \in \Sigma$ is defined to be $\int_A s \, d\mu = \sum_{i=1}^n x_i \mu(A \cap A_i)$. A function $f \colon \Omega \to X$ is strongly measurable if it is the $\mu$-almost everywhere limit of a sequence of simple functions.

Definition 2.3. A strongly measurable function $f \colon \Omega \to X$ is Bochner integrable if there is a sequence $(s_n)$ of simple functions such that \[\int_\Omega \|f - s_n\| \, d\mu \to 0.\] In this case, the Bochner integral of $f$ over $A \in \Sigma$ is defined to be \[\int_A f \, d\mu = \lim_n \int_A s_n \, d\mu.\] The integral does not depend on the choice of $(s_n)$.

A theorem of Bochner states that a strongly measurable function $f$ is Bochner integrable if and only if $\int_\Omega \|f\| \, d\mu < \infty$. We denote by $L_1(\mu; X)$ the space of (equivalence classes up to $\mu$-almost everywhere equality of) Bochner integrable functions. This is a Banach space when equipped with the norm \[\|f\|_1 = \int_\Omega \|f\| \, d\mu.\] Every $f \in L_1(\mu; X)$ produces a vector measure $F(A) = \int_A f \, d\mu$, which is $\mu$-continuous and of bounded variation; in fact, $|F|(A) = \int_A \|f\| \, d\mu$. The RNP asks for the converse to hold.

Definition 2.4. A Banach space $X$ has the Radon–Nikodým property (RNP) if for every finite measure space $(\Omega, \Sigma, \mu)$ and every $\mu$-continuous vector measure $F \colon \Sigma \to X$ of bounded variation there exists $f \in L_1(\mu; X)$ such that \[F(A) = \int_A f \, d\mu \quad \text{for every } A \in \Sigma.\]

Remark. In this language, the Radon–Nikodým theorem simply says that $\R$ has the RNP.

There is also a characterization of the RNP in terms of vector-valued martingales. We will mainly use the geometric characterization of the RNP, which is explained below, but since vector-valued conditional expectations and martingales will appear in the proof of the main theorem, this seems like a good excuse to introduce them.

Proposition 2.5. Let $\mathcal{A} \subseteq \Sigma$ be a sub-$\sigma$-algebra. For every $f \in L_1(\mu; X)$ there exists a unique (up to $\mu$-almost everywhere equality) $\mathcal{A}$-strongly measurable function $\mathbb{E}(f \mid \mathcal{A}) \in L_1(\mu; X)$ such that \[\int_A \mathbb{E}(f \mid \mathcal{A}) \, d\mu = \int_A f \, d\mu \quad \text{for every } A \in \mathcal{A}.\] Moreover, \[\|\mathbb{E}(f \mid \mathcal{A})\| \leq \mathbb{E}(\|f\| \mid \mathcal{A}) \quad \text{$\mu$-almost everywhere}.\]

Denote by $L_\infty(\mu; X)$ the space of (equivalence classes of) strongly measurable functions $f \colon \Omega \to X$ such that \[\|f\|_\infty = \operatorname{ess\,sup}_{\omega \in \Omega} \|f(\omega)\| < \infty.\] As in the scalar case, $L_\infty(\mu; X)$ is a Banach space when equipped with this norm. The following fact will be used in the proof of the main theorem.

Corollary 2.6. For every $f \in L_\infty(\mu; X)$, we have $\|\mathbb{E}(f \mid \mathcal{A})\|_\infty \leq \|f\|_\infty$.

Definition 2.7. A filtration is an increasing sequence $(\Sigma_n)$ of sub-$\sigma$-algebras of $\Sigma$. A sequence $(f_n)$ in $L_1(\mu; X)$ is a martingale with respect to $(\Sigma_n)$ if, for every $n \in \N$, $f_n$ is $\Sigma_n$-strongly measurable and \[\mathbb{E}(f_{n+1} \mid \Sigma_n) = f_n.\] The martingale is uniformly bounded if $\sup_n \|f_n\|_\infty < \infty$.

The link between martingales and the RNP is already apparent in the following example. Let $(\pi_n)$ be a sequence of finite partitions of $\Omega$ into sets of positive measure such that $\pi_{n+1}$ refines $\pi_n$, and let $\Sigma_n$ be the (finite) $\sigma$-algebra generated by $\pi_n$. Given a vector measure $F \colon \Sigma \to X$, define \[f_n = \sum_{A \in \pi_n} \frac{F(A)}{\mu(A)} \chi_A.\] For every $A \in \pi_n$, additivity of $F$ gives \[\int_A f_{n+1} \, d\mu = \sum_{\substack{B \in \pi_{n+1} \\ B \subseteq A}} F(B) = F(A) = \int_A f_n \, d\mu,\] so $(f_n)$ is a martingale with respect to $(\Sigma_n)$. If moreover $\|F(A)\| \leq C \mu(A)$ for every $A \in \Sigma$, then the martingale is uniformly bounded by $C$. Whenever $F$ has a density $f$, we have $f_n = \mathbb{E}(f \mid \Sigma_n)$, and the martingale converges to $f$ if $\Sigma$ is generated by $\bigcup_n \Sigma_n$. Conversely, the limit of $(f_n)$, if it exists, is the natural candidate for a density of $F$. This idea leads to the martingale characterization of the RNP.

Theorem 2.8 (Chatterji [7]). A Banach space $X$ has the RNP if and only if, for every finite measure space $(\Omega, \Sigma, \mu)$ and every filtration $(\Sigma_n)$, every uniformly bounded $X$-valued martingale with respect to $(\Sigma_n)$ converges $\mu$-almost everywhere and in $L_1(\mu; X)$.

We have yet to give examples of spaces with or without the RNP. However, we will wait to do so until we have one further characterization of the RNP, in this case a geometric one. It is remarkable that the RNP, which in principle has to do with measure-theoretic and probabilistic notions, also admits a clean geometric characterization in terms of dentable sets.

From now on, we will stick to real Banach spaces. Given $x \in X$ and $\varepsilon > 0$, we denote by $B(x, \varepsilon)$ the open ball centered at $x$ with radius $\varepsilon$, and given $A \subseteq X$, we denote by $\operatorname{co}(A)$ its convex hull, and by $\overline{\mathrm{co}}(A)$ its closed convex hull.

Definition 2.9. A nonempty bounded set $A \subseteq X$ is dentable if, for every $\varepsilon > 0$, there exists $x \in A$ such that \[x \notin \overline{\mathrm{co}}\big(A \setminus B(x, \varepsilon)\big).\]

In other words, $A$ is dentable if, for every $\varepsilon > 0$, one can find a point of $A$ that cannot be reached with convex combinations (and limits of them) of the points of $A$ that are at distance at least $\varepsilon$ from it.

Remark. Dentability is also commonly formulated in terms of slices. Let $A \subseteq X$ be nonempty and bounded, and let $f \in X^*$. We write \[M(A, f) = \sup_{x \in A} f(x),\] which is finite because $A$ is bounded. For $\alpha > 0$, the slice of $A$ determined by $f$ and $\alpha$ is defined to be \[S(A, f, \alpha) = \{ x \in A : f(x) > M(A, f) - \alpha \}.\] By definition of the supremum, every slice is nonempty. Geometrically, a slice is the part of $A$ that is cut off by a hyperplane parallel to a supporting hyperplane of $A$.

The set $A$ is dentable if and only if it has slices of arbitrarily small diameter, that is, if for every $\varepsilon > 0$ there exist $f \in X^*$ and $\alpha > 0$ such that \[\operatorname{diam} S(A, f, \alpha) < \varepsilon.\] Indeed, suppose first that $\operatorname{diam} S(A, f, \alpha) < \varepsilon$, and pick any $x \in S(A, f, \alpha)$. Every point of the slice is at distance less than $\varepsilon$ from $x$, so \[A \setminus B(x, \varepsilon) \subseteq \{ y \in X : f(y) \leq M(A, f) - \alpha \}.\] The right-hand side is closed and convex, so it also contains $\overline{\mathrm{co}}(A \setminus B(x, \varepsilon))$. Since $f(x) > M(A, f) - \alpha$, we conclude that $x \notin \overline{\mathrm{co}}(A \setminus B(x, \varepsilon))$.

Conversely, let $\varepsilon > 0$ and let $x \in A$ be such that $x \notin C = \overline{\mathrm{co}}(A \setminus B(x, \varepsilon))$. If $C$ is empty, then $A \subseteq B(x, \varepsilon)$, and every slice of $A$ has diameter at most $2\varepsilon$. Otherwise, by the Hahn–Banach separation theorem there exists $f \in X^*$ such that \[f(x) > \beta = \sup_{y \in C} f(y).\] Let $\alpha = M(A, f) - \beta$, which is positive because $M(A, f) \geq f(x) > \beta$. Then $S(A, f, \alpha) = \{ y \in A : f(y) > \beta \}$ is disjoint from $A \setminus B(x, \varepsilon)$, that is, $S(A, f, \alpha) \subseteq B(x, \varepsilon)$, and so $\operatorname{diam} S(A, f, \alpha) \leq 2\varepsilon$. Since $\varepsilon > 0$ is arbitrary, $A$ has slices of arbitrarily small diameter.

$\alpha$

$f$ = () $M(A,f)$ = $\alpha$ = $\operatorname{diam} S(A,f,\alpha)$ =

Figure 1. $A$ is a set in the plane, chosen below; with draw you can trace your own. The blue region is the slice $S(A,f,\alpha)$: the points of $A$ where $f$ is larger than $M(A,f) - \alpha$, and $M(A,f)$ is reached at the marked point. The arrow is $f$: drag it to change $f$, and use the slider to change $\alpha$. On the right, the values of $f$ are drawn as heights, and the slice is the part that rises above the blue plane.

The same idea can be localized at a single point of $A$.

Definition 2.10. Let $A \subseteq X$ be nonempty and bounded. A point $x \in A$ is a dentable point of $A$ (we also say that $x$ is dentable in $A$) if, for every $\varepsilon > 0$, \[x \notin \overline{\mathrm{co}}\big(A \setminus B(x, \varepsilon)\big).\]

Clearly, a set with a dentable point is dentable. The converse is false: the open unit ball of $\R^2$ is dentable, but none of its points is a dentable point, since every $x$ in it is the midpoint of two points of the ball at distance at least $\varepsilon$ from $x$, for $\varepsilon > 0$ small enough.

Remark. The argument of the previous remark shows that $x \in A$ is a dentable point of $A$ if and only if, for every $\varepsilon > 0$, there exist $f \in X^*$ and $\alpha > 0$ such that \[x \in S(A, f, \alpha) \quad \text{and} \quad \operatorname{diam} S(A, f, \alpha) < \varepsilon.\]

Theorem 2.11 ([8, Chapter 6]). A Banach space has the Radon–Nikodým property if and only if every bounded subset is dentable.

Corollary 2.12 ([8, Chapter 6, Corollary 3]). For a Banach space $X$, the following are equivalent:

(i)
$X$ has the RNP.
(ii)
Every closed subspace of $X$ has the RNP.
(iii)
Every separable closed subspace of $X$ has the RNP.

Example 2.13. The unit ball $B_{L_1[0,1]}$ is not dentable. Indeed, we will see that every $f \in B_{L_1[0,1]}$ is the average of two elements $g, h \in B_{L_1[0,1]}$ with $\|g - f\|_1 = \|h - f\|_1 = 1$. In particular, for every $f \in B_{L_1[0,1]}$, \[f \in \operatorname{co}\big(B_{L_1[0,1]} \setminus B(f, 1)\big),\] so $B_{L_1[0,1]}$ is not dentable. If $f = 0$, take $g = \chi_{[0,1]}$ and $h = -\chi_{[0,1]}$. Otherwise, the function $t \mapsto \int_0^t |f|$ is continuous, so by the intermediate value theorem there exists $c \in [0,1]$ such that \[\int_0^c |f| = \int_c^1 |f| = \frac{\|f\|_1}{2}.\] Let $\lambda = 1/\|f\|_1 \geq 1$ and $u = \lambda f (\chi_{[0,c]} - \chi_{(c,1]})$, and define $g = f + u$ and $h = f - u$. Clearly $f = (g + h)/2$ and $\|g - f\|_1 = \|h - f\|_1 = \|u\|_1 = \lambda \|f\|_1 = 1$. Moreover, \[g = (1 + \lambda) f \chi_{[0,c]} + (1 - \lambda) f \chi_{(c,1]},\] so that \[\|g\|_1 = (1 + \lambda) \frac{\|f\|_1}{2} + (\lambda - 1) \frac{\|f\|_1}{2} = \lambda \|f\|_1 = 1,\] and similarly $\|h\|_1 = 1$. By the previous theorem, $L_1[0,1]$ does not have the RNP.

Example 2.14. The unit ball $B_{c_0}$ is not dentable. Let $x = (x_n) \in B_{c_0}$. Since $x_n \to 0$, there is $N \in \N$ such that $|x_N| \leq 1/2$. Denote by $e_N$ the $N$-th unit vector, and let $y = x + \frac{1}{2} e_N$ and $z = x - \frac{1}{2} e_N$. Only the $N$-th coordinate changes, and $|x_N \pm 1/2| \leq 1$, so $y, z \in B_{c_0}$. Moreover, $\|y - x\|_\infty = \|z - x\|_\infty = 1/2$ and $x = (y + z)/2$. Hence, for every $x \in B_{c_0}$, \[x \in \operatorname{co}\big(B_{c_0} \setminus B(x, 1/2)\big),\] so $B_{c_0}$ is not dentable. By the previous theorem, $c_0$ does not have the RNP.

What about spaces with the RNP? The following theorem provides plenty.

Theorem 2.15 ([8, Chapter 6]).

(i)
Every separable dual space has the RNP.
(ii)
Every reflexive space has the RNP.

In Example 2.13 and Example 2.14, we showed that these spaces fail the RNP by checking that their unit balls are not dentable. In both examples, moreover, the unit ball has no extreme points. However, this is not necessary for the unit ball to fail to be dentable.

Example 2.16. Let $\Delta$ be the Cantor set and let $B$ be the unit ball of $C(\Delta)$. This unit ball has extreme points: every $f \in B$ with $|f|=\chi _\Delta$ is an extreme point.

However, every slice of $B$ has diameter $2$. By the Riesz representation theorem, every $\varphi \in C(\Delta)^*$ is integration against a regular signed Borel measure $\mu$ on $\Delta$, with $\|\varphi\| = |\mu|(\Delta)$, so that $M(B, \varphi) = |\mu|(\Delta)$. Let $\alpha > 0$ and choose $f \in B$ with $\int f \, d\mu > |\mu|(\Delta) - \alpha/4$. For every $n \in \N$, the Cantor set can be partitioned into $2^n$ nonempty clopen pieces, whose $|\mu|$-masses must add up to $|\mu|(\Delta)$. Taking $n$ large enough, some piece $U$ satisfies $|\mu|(U) < \alpha/4$. Now set \[g_\pm = f \chi_{\Delta \setminus U} \pm \chi_U.\] These functions are continuous (because $U$ is clopen) and lie in $B$, and \[\Big| \int g_\pm \, d\mu - \int f \, d\mu \Big| = \Big| \int_U (\pm 1 - f) \, d\mu \Big| \leq 2 |\mu|(U) < \frac{\alpha}{2},\] so both are in $S(B, \varphi, \alpha)$. But $g_+ - g_- = 2 \chi_U$, so $\|g_+ - g_-\|_\infty = 2$. In particular, $B$ is not dentable, and $C(\Delta)$ does not have the RNP.

One may still have the feeling that dentability is somehow related to extreme points. Bourgain himself had established the following close relation between the RNP and extreme points in an earlier paper.

Theorem 2.17 (Bourgain [5, Corollary 4]). A Banach space $X$ has the RNP if and only if every nonempty bounded closed convex set of $X$ contains an extreme point of its weak$^*$-closure in $X^{**}$.

This certainly resonates with the Krein–Milman property.

3. The Krein–Milman property

Let us start by recalling the notion of extreme point and the classical Krein–Milman theorem, which will motivate the introduction of the Krein–Milman property.

Definition 3.1. Let $C$ be a nonempty closed convex subset of a topological vector space. A subset $D$ of $C$ is extremal in $C$ if $D$ is itself a nonempty closed convex set and it has the following property: if $x,y \in C$ and $tx+(1-t)y\in D$, for some $t \in (0,1)$, then $x,y \in D$. An extreme point of $C$ is a point $x$ in $C$ such that $\{x\}$ is an extremal subset of $C$. The set of extreme points of $C$ is denoted by $\ext(C)$.

Remark. Let $C$ be a nonempty closed convex subset of a Hausdorff topological vector space, and let $x \in C$. Then $x$ is an extreme point of $C$ if and only if whenever $x_1,x_2 \in C$ and $x=(x_1+x_2)/2$, it follows that $x_1=x_2=x$.

Click a point $x$ of $C$.

Figure 2. Extreme points of convex sets $C$ in the plane and in space, marked in blue. Click a point $x$ of $C$ (in the plane, you can also drag it): if $x$ is not extreme, it appears as the midpoint of two distinct points $y$ and $z$ of $C$, as in the previous remark. Choose draw to trace your own set: $C$ is then the closed convex hull of the curve. Drag the solids to turn them.

Theorem 3.2 (Krein–Milman). Let $C$ be a nonempty compact convex subset of a Hausdorff locally convex topological vector space. Then $C$ is the closed convex hull of its set of extreme points, that is, $C=\co(\ext(C))$.

This theorem has many important consequences. For example, Example 2.14 shows that the unit ball of $c_0$ does not have extreme points, hence it cannot be compact in any locally convex Hausdorff linear topology on $c_0$; in particular, $c_0$ is not linearly isometric to the dual of any Banach space (for in this case it would have the weak$^*$ topology, a locally convex Hausdorff linear topology for which the unit ball is compact).

Every compact subset of a Hausdorff topological vector space is both bounded and closed. Hence it is natural to wonder when a stronger version of the Krein–Milman theorem, concerning closed bounded subsets, holds.

Definition 3.3. A Banach space $X$ has the Krein–Milman property (KMP) if every nonempty closed bounded convex subset $C$ of $X$ is the closed convex hull of its set of extreme points.

In particular, if $X$ has the KMP, then every nonempty closed bounded convex subset of $X$ has at least one extreme point. Surprisingly, this apparently weaker condition is equivalent to the KMP, as we will see below.

Since being an extreme point of $C$ only depends on $C$ and not on the ambient space, the KMP passes to closed subspaces. Moreover, a linear isomorphism between Banach spaces maps nonempty closed bounded convex sets onto nonempty closed bounded convex sets, extreme points onto extreme points, and closed convex hulls onto closed convex hulls, so the KMP is invariant under isomorphisms. Consequently, if a Banach space contains a closed subspace isomorphic to a space failing the KMP, then it fails the KMP as well. Example 2.13 and Example 2.14 show that the unit balls of $L_1[0,1]$ and $c_0$ have no extreme points, so neither $L_1[0,1]$ nor $c_0$ has the KMP. In particular, no Banach space with the KMP contains a subspace isomorphic to $c_0$. Moreover, $C(\Delta)$ does not have the KMP either, because it contains an isometric copy of $c_0$.

If you are looking for examples of spaces with the Krein–Milman property, the following result provides plenty.

Theorem 3.4 (Lindenstrauss [14]). Every Banach space with the Radon–Nikodým property has the Krein–Milman property.

The proof of this theorem is elegant and direct, and employs techniques that will be useful for our main theorem, so it is worth presenting it in detail. Its most important tool is a variation of the Bishop–Phelps theorem. Let us start by recalling this classical result.

Definition 3.5. Let $X$ be a Banach space and let $C$ be a nonempty closed convex subset of $X$. A functional $f \in X^*$, $f \neq 0$, is a support functional of $C$ if it attains its supremum on $C$, that is, if there exists $x_0 \in C$ such that \[f(x_0) = \sup_{x \in C} f(x).\] In this case, $x_0$ is a support point of $C$, and we say that $f$ supports $C$ at $x_0$.

When $C$ is weakly compact, every functional attains its supremum on $C$, so every nonzero functional is a support functional. In general, there are functionals that do not attain their supremum on $C$, but the Bishop–Phelps theorem guarantees that we can always perturb them slightly so that they do.

Theorem 3.6 (Bishop–Phelps [3]). Let $C$ be a nonempty closed bounded convex subset of a Banach space $X$. Then the set of support functionals of $C$ is norm-dense in $X^*$.

We will use the Bishop–Phelps theorem in the following form, which strengthens the Hahn–Banach separation theorem: a functional separating a point of $C$ from a smaller closed convex set can be chosen to attain its supremum on $C$.

Corollary 3.7. Let $X$ be a Banach space and let $B \subseteq C$ be nonempty closed convex subsets of $X$ such that $C$ is bounded and $B \neq C$. Then there exist $f \in X^*$ and $c' \in C$ such that \[\sup_{b \in B} f(b) < \max_{c \in C} f(c) = f(c').\]

Proof. Let $c \in C \setminus B$. By the Hahn–Banach separation theorem, there exists $h \in X^*$ such that $h(c) > \sup_{b \in B} h(b)$. Let \[\eta = h(c) - \sup_{b \in B} h(b) > 0.\] Let $r = \sup_{x \in C} \|x\|$, which is finite because $C$ is bounded, and positive because $C$ contains at least two points. By the Bishop–Phelps theorem, there exists a support functional $f$ of $C$ such that $\|f - h\| < \eta/(3r)$. Let $c' \in C$ be such that $f(c') = \max_{c \in C} f(c)$. Since $|f(x) - h(x)| < \eta/3$ for every $x \in C$, we get, for every $b \in B$, \[f(b) < h(b) + \frac{\eta}{3} \leq h(c) - \frac{2\eta}{3} < f(c) - \frac{\eta}{3}.\] Taking the supremum over $b \in B$, we get $\sup_{b \in B} f(b) \leq f(c) - \eta/3 < f(c')$.□

Now we can proceed to the proof of Theorem 3.4. First, a preliminary proposition that is of interest in its own right.

Proposition 3.8. Suppose that every nonempty closed bounded convex subset of a Banach space $X$ has an extreme point. Then $X$ possesses the Krein–Milman property.

Proof. Let $C$ be a nonempty closed bounded convex subset of $X$, and let \[E = \co(\ext(C)).\] By assumption, $C$ has an extreme point, so $E$ is a nonempty closed convex subset of $C$. Suppose, for a contradiction, that $E \neq C$. By Corollary 3.7, there exist $f \in X^*$ and $c' \in C$ such that \[f(e) < \max_{c \in C} f(c) = f(c') \quad \text{for every } e \in E.\] Let \[K = \{ x \in C : f(x) = f(c') \}.\] By the choice of $f$ and $c'$, the set $K$ is nonempty, closed, bounded and convex, so by assumption it has an extreme point $x$.

We claim that $x$ is also an extreme point of $C$. First, $K$ is extremal in $C$: if $y, z \in C$ and $ty + (1-t)z \in K$ for some $t \in (0,1)$, then \[t f(y) + (1-t) f(z) = f(c'),\] and since $f(y), f(z) \leq f(c')$, this forces $f(y) = f(z) = f(c')$, that is, $y, z \in K$. Now let $y, z \in C$ and $t \in (0,1)$ be such that $ty + (1-t)z = x$. Since $x \in K$ and $K$ is extremal in $C$, we get $y, z \in K$, and since $x$ is an extreme point of $K$, we get $y = z = x$. This proves the claim.

Therefore $x \in \ext(C) \subseteq E$, so $f(x) < f(c')$, which contradicts $x \in K$. Hence $E = C$, and $X$ has the KMP.□

Remark. Hence the KMP is equivalent to the property that every nonempty closed bounded convex subset has an extreme point.

Proof of Theorem 3.4. By Proposition 3.8, it suffices to show that every nonempty closed bounded convex subset $C$ of $X$ has an extreme point. Since $X$ has the RNP, every nonempty bounded subset of $X$ is dentable; in particular, so are $C$ and each of its nonempty subsets.

Let $\varepsilon > 0$. We construct inductively a decreasing sequence $(C_n)$ of nonempty closed bounded convex subsets of $C$, together with functionals $f_n \in X^*$, as follows. Let $C_0 = C$, and suppose that $C_{n-1}$ has already been constructed for some $n \in \N$. Since $C_{n-1}$ is dentable, there exists $x_n \in C_{n-1}$ such that \[x_n \notin \co\big(C_{n-1} \setminus B(x_n, \varepsilon / 2^n)\big) = D_n.\] If $D_n$ is empty, then $C_{n-1} \subseteq B(x_n, \varepsilon / 2^n)$, and we let $C_n = C_{n-1}$ and $f_n = 0$. Otherwise, $D_n$ is a nonempty closed convex subset of $C_{n-1}$ different from $C_{n-1}$, so by Corollary 3.7 there exist $f_n \in X^*$ and $c_n \in C_{n-1}$ such that \[f_n(d) < \max_{c \in C_{n-1}} f_n(c) = f_n(c_n) \quad \text{for every } d \in D_n.\] In this case, let \[C_n = \{ c \in C_{n-1} : f_n(c) = f_n(c_n) \}.\] In both cases, $C_n$ is a nonempty closed bounded convex subset of $C_{n-1}$ such that \[f_n(c) = \max_{C_{n-1}} f_n \quad \text{for every } c \in C_n.\] Moreover, $C_n$ is disjoint from $D_n$, and in particular from $C_{n-1} \setminus B(x_n, \varepsilon / 2^n)$, so that \[\operatorname{diam} C_n \leq \operatorname{diam} B(x_n, \varepsilon /2^{n})=\frac{\varepsilon}{2^{n-1}}.\]

Since $(C_n)$ is a decreasing sequence of nonempty closed subsets of the Banach space $X$ whose diameters tend to zero, there exists $x \in X$ such that \[\bigcap_{n \in \N} C_n = \{x\}.\] We claim that $x$ is an extreme point of $C$. Let $y, z \in C$ be such that $x = (y + z)/2$. We show by induction that $y, z \in C_n$ for every $n \geq 0$. This is clear for $n = 0$. If $y, z \in C_{n-1}$, then, since $x \in C_n$, \[\max_{C_{n-1}} f_n = f_n(x) = \frac{f_n(y) + f_n(z)}{2},\] and since $f_n(y), f_n(z) \leq \max_{C_{n-1}} f_n$, this forces $f_n(y) = f_n(z) = \max_{C_{n-1}} f_n$, that is, $y, z \in C_n$. Hence $y, z \in \bigcap_n C_n = \{x\}$, so $y = z = x$, and $x$ is an extreme point of $C$.□

All the examples of spaces with the RNP in the previous section are also examples of spaces with the KMP. Can we find other examples? According to F. Albiac and N. J. Kalton [1, p. 131]:

The converse [to Theorem 3.4] remains the major open problem in this area […] It is probably fair to say that the subject has received relatively little attention since the 1980s, and some really new ideas seem to be necessary to make further progress.

There are, however, important partial converses.

Theorem 3.9 (Huff–Morris [12]). A dual Banach space $X^*$ has the RNP if and only if it has the KMP.

Theorem 3.10 (Schachermayer [19]). Let $X$ be a Banach space that is isomorphic to its square $X \oplus X$. Then $X$ has the RNP if and only if it has the KMP.

The goal of the main paper [6] is to add Banach lattices to the list of spaces in which the KMP implies the RNP.

4. Banach lattices

Recall the definition of a Banach lattice.

Definition 4.1. A Banach lattice is a real Banach space $E$ equipped with a partial order $\leq$ such that, for every $x,y \in E$,

(i)
$x \leq y$ implies $x + z \leq y + z$ for every $z \in E$;
(ii)
$x \leq y$ implies $\lambda x \leq \lambda y$ for every $\lambda \geq 0$;
(iii)
the supremum $x \vee y$ and the infimum $x \wedge y$ exist;
(iv)
if $|x|\le |y|$, where $|x|=x\vee (-x)$, then $\|x\|\le \|y\|$.

Many classical Banach spaces are Banach lattices in a natural way. For instance, this is the case for spaces of continuous functions, with the pointwise order, and for $L_p$ spaces, with the order defined pointwise almost everywhere. In fact, every identity involving the order and the linear operations of a Banach lattice holds as long as it holds for real numbers [16, Theorem 2.1.20]. This is a remarkable fact of the theory that is very useful in practice.

There is a well-developed theory of Banach lattices [2, 15, 16, 20]. Some of its most beautiful and deepest results relate order-theoretic properties of a Banach lattice to purely linear properties coming from Banach space theory. They show that the order structure is not merely built on top of the linear structure, but that it also has an influence on the linear structure itself. Sometimes we even find statements that make sense in every Banach space, because they only involve linear properties, and that can be proved for Banach lattices but are false, or not known to be true, for general Banach spaces. The main theorem we want to explore, namely that the RNP and the KMP are equivalent in Banach lattices, is a result of this kind.

We will now present some preliminary facts about Banach lattices that will be used in the main construction. Not surprisingly, along the way, we will find several instances of results that relate linear and order properties. We will start with the notion of order continuity, which captures when convergence in order implies convergence in norm.

Definition 4.2. A Banach lattice $E$ is order continuous (or has an order continuous norm) if, whenever $(x_\alpha)$ is a decreasing net in $E$ with $\inf_\alpha x_\alpha =0$, $\lim_\alpha \|x_\alpha \|=0$.

For example, the spaces $c_0$ and $L_p(\mu )$, for any measure $\mu$ and $1\le p<\infty$, are order continuous; however, $\ell_\infty$ is not.

It turns out that being order continuous admits several equivalent characterizations, including a purely linear one. Remember that a set in a Banach lattice is said to be order bounded if it is contained in an order interval.

Theorem 4.3 (Nakano). For a Banach lattice $E$, the following are equivalent:

(i)
$E$ is order continuous.
(ii)
Every order interval of $E$ is weakly compact.
(iii)
Every monotone order bounded net in $E_+$ is norm convergent (to its supremum/infimum).
(iv)
Every monotone order bounded sequence in $E_+$ is norm convergent (to its supremum/infimum).
(v)
The canonical embedding of $E$ into its bidual $E^{* *}$ embeds $E$ as an order ideal in $E^{* *}$.

There is also a local version of the previous result that will be useful in the main proof.

Theorem 4.4 ([16, Theorem 3.4.4]). For a positive vector $u$ in a Banach lattice $E$, the following are equivalent:

(i)
Every monotone net in $[0,u]$ is norm convergent (to its supremum/infimum).
(ii)
$[0,u]$ is weakly compact.

Condition (iv) in Theorem 4.3 asks increasing sequences to converge when they are order bounded. What happens if we only ask them to be norm bounded? In $L_1(\mu )$, for instance, such sequences do converge, by the Monotone Convergence Theorem. This suggests the following definition.

Definition 4.5. A Banach lattice $E$ is a KB-space (after Kantorovich and Banach) if every increasing norm bounded sequence in $E_+$ is norm convergent.

Since an order bounded sequence $0 \leq x_n \leq x$ is norm bounded by $\|x\|$, Theorem 4.3 shows that every KB-space is order continuous. The converse fails, as $c_0$ shows.

As with order continuity, being a KB-space also admits a purely linear characterization.

Theorem 4.6 ([2, Theorem 4.60]). For a Banach lattice $E$, the following are equivalent:

(i)
$E$ is a KB-space.
(ii)
$c_0$ is not lattice embeddable in $E$.
(iii)
$c_0$ is not embeddable in $E$.

In particular, if a Banach lattice contains a copy of $c_0$, then it also contains a copy of $c_0$ that respects the order structure.

Going back to the RNP, Ghoussoub and Talagrand provided yet another characterization of this property, this time using the order structure of Banach lattices. Recall that, in a Banach space, a set is dentable if it has slices of arbitrarily small diameter. In a Banach lattice, there is another natural way of measuring how “small” a positive element is, which uses the order. An element $u \in E_+$ is a quasi-interior point of $E$ if \[\lim_{n \to \infty} \|x - x \wedge nu\| = 0 \quad \text{for every } x \in E_+.\] If $u$ is a quasi-interior point and $x \in E_+$ satisfies $x \wedge u = 0$, then $x \wedge nu \leq n (x \wedge u) = 0$ for every $n \in \N$, so $x = 0$. Thus, for $x \in E_+$, the quantity $\|x \wedge u\|$, which measures the part of $x$ that lies below $u$, is a weak way of measuring how far $x$ is from $0$. It may be much smaller than $\|x\|$: in $L_1[0,1]$ with $u = \chi_{[0,1]}$, the “spike” $x = n \chi_{[0, 1/n]}$ has $\|x\|_1 = 1$ but $\|x \wedge u\|_1 = 1/n$.

A set of positive elements is order dentable if it has a slice that stays away from $0$ in this weak sense. Equivalently (see the remark below), it cannot be recovered, by taking closed convex hulls, from its elements that are arbitrarily close to $0$ in this weak sense, such as the spikes above. Ghoussoub and Talagrand were motivated by martingales: if a bounded closed convex set is not order dentable, one can build a martingale with values in it that is hopelessly divergent, and so the RNP fails [10, Lemma 2].

Definition 4.7. Let $E$ be a Banach lattice with a quasi-interior point $u$, and for $n \in \N$ let \[H_n = \{ x \in E_+ : \|x \wedge u\| \geq 1/n \}.\] A nonempty bounded set $A \subseteq E_+$ is order dentable if there exist $f \in E^*$, $\alpha > 0$ and $n \in \N$ such that \[S(A, f, \alpha) \subseteq H_n.\] A Banach lattice $E$ is order dentable if, for every closed sublattice $F$ of $E$ with a quasi-interior point, every closed bounded convex set $C \subseteq F_+$ with $C \neq \{0\}$ is order dentable in $F$.

Remark. Order dentability can also be formulated in terms of closed convex hulls, which is the original formulation of [10]. Let $A \subseteq E_+$ be nonempty and bounded. Then $A$ is order dentable if and only if there exists $n \in \N$ such that \[A \not\subseteq \overline{\mathrm{co}}(A \setminus H_n),\] that is, if $A$ cannot be recovered, by taking closed convex hulls, from its elements $y$ with $\|y \wedge u\| < 1/n$. When $A$ is closed and convex, since the sets $A \setminus H_n$ decrease with $n$, this is the same as saying that \[A \neq \bigcap_{n \in \N} \overline{\mathrm{co}}(A \setminus H_n).\] The proof is the same as in the remark on slices in Section 2. Suppose first that $S(A, f, \alpha) \subseteq H_n$. Then \[A \setminus H_n \subseteq A \setminus S(A, f, \alpha) \subseteq \{ y \in E : f(y) \leq M(A, f) - \alpha \}.\] The right-hand side is closed and convex, so it also contains $\overline{\mathrm{co}}(A \setminus H_n)$, and therefore no point of the (nonempty) slice $S(A, f, \alpha)$ belongs to $\overline{\mathrm{co}}(A \setminus H_n)$.

Conversely, let $n \in \N$ and $x \in A$ be such that $x \notin C = \overline{\mathrm{co}}(A \setminus H_n)$. If $C$ is empty, then $A \subseteq H_n$, and every slice of $A$ is contained in $H_n$. Otherwise, by the Hahn–Banach separation theorem there exists $f \in E^*$ such that \[f(x) > \beta = \sup_{y \in C} f(y).\] Let $\alpha = M(A, f) - \beta$, which is positive because $M(A, f) \geq f(x) > \beta$. Then $S(A, f, \alpha) = \{ y \in A : f(y) > \beta \}$ is disjoint from $A \setminus H_n$, that is, $S(A, f, \alpha) \subseteq H_n$.

The notion of order dentable set does not depend on the choice of the quasi-interior point $u$ [10, Lemma 1]. Moreover, it is enough to check the definition of order dentable Banach lattice for separable closed sublattices [10, Proposition 1].

The space $c_0$ is order dentable [10, p. 218], even though it fails the RNP. This explains the extra condition in the following theorem.

Theorem 4.8 (Ghoussoub–Talagrand [10, Theorem 2]). For a Banach lattice $E$, the following are equivalent:

(i)
$E$ has the RNP.
(ii)
$E$ is order dentable and $c_0$ is not lattice embeddable in $E$.

By Theorem 4.6, condition (ii) can be equivalently stated as: $E$ is an order dentable KB-space.

After these ((very) long) preliminaries, we are now ready to present the construction of [6].

5. The main theorem

In [6], Bourgain and Talagrand provide two proofs of the following theorem.

Theorem 5.1. In a Banach lattice, the Radon–Nikodým property and the Krein–Milman property are equivalent.

Remember that the RNP always implies the KMP (Theorem 3.4). Hence the content of the theorem lies in the reverse implication. The first proof builds on Theorem 2.17.

Proof 1. Suppose that $E$ is a Banach lattice with the Krein–Milman property. Since $E$ has the KMP, it cannot contain $c_0$. Hence, according to Theorem 4.6, it must be a KB-space, and therefore $E$ sits as an order ideal in its bidual (Theorem 4.3). According to Theorem 2.17, if $E$ fails the RNP, then there exists a nonempty bounded closed convex set $C$ of $E$ that does not contain any extreme point of its weak$^*$ closure $\overline{C}^{w^*}$ in $E^{* *}$. In fact, it follows from [5, 10] that the set $C$ may be taken to be in $E_+$. This can be justified by noting that positive martingales suffice to test the RNP: if every $L_1$-bounded martingale with values in $E_+$ converges almost everywhere, then $E$ has the RNP [10, Lemma 3]. The reason is that, in a Banach lattice that does not contain $c_0$, every $L_1$-bounded martingale is the difference of two positive $L_1$-bounded martingales. One then uses the construction of [5] in $E_+$. However, these details would require a whole paper of their own, and since the goal of [6] is to give a simplified proof using an order-theoretic approach, there is no point in delving deeper.

Granting this, it is not difficult to finish the proof. The set $C$, being closed and convex, is weakly closed by Mazur’s theorem. Since the weak topology of $E$ is the same, by definition, as the weak$^*$ topology of $E^{* *}$ when restricted to $E$, it follows that $\overline{C}^{w^{*}}\cap E=C$. Now the fact that the elements of $C$ are positive implies $\ext(C)\subseteq \ext(\overline{C}^{w^{*}})$: if $x \in \ext(C)$, and $x=(y+z)/2$ for some $y,z \in \overline{C}^{w^{*}}\subseteq E^{* *}_+$, then $0\le y\le 2x$ and $0\le z\le 2x$. Since $E$ is an order ideal in $E^{* *}$, we get $y,z \in E\cap \overline{C}^{w^{*}}=C$, and therefore $x=y=z$. Thus $\ext(C)\subseteq \ext(\overline{C}^{w^{*}})\cap E$, which is empty by assumption. This contradicts the Krein–Milman property.□

As we have seen, this proof relies heavily on the construction of [5]. However, according to the authors,

The techniques of [5] are, however, rather delicate, and they simplify notably if one uses the order techniques introduced in [11], although the principle of the construction is unchanged. It therefore seemed to us that this approach has an interest of its own.

This new construction requires five preparatory lemmas, several of which may be of independent interest. Some of them deal with weakly compact intervals. This is natural: we will be working with Banach lattices that have the KMP, hence do not contain $c_0$; in particular, they have weakly compact intervals. The first lemma has to do with the geometry of order continuous Banach lattices.

Lemma 5.2 (Ghoussoub [11]). Let $E$ be a Banach lattice, and let $u \in E_+$ be such that $[0,u]$ is weakly compact. Then $0$ is a dentable point of $[0,u]$.

Proof. From the original proof: “We shall take pains to give an elementary one.” Assume $u\neq 0$, for otherwise the result is trivial. The proof has the following steps:

(i)
First, show that there exists $v \in [0,u]$ different from zero and such that $0$ is dentable in $[0,v]$.
(ii)
Second, show that the set $A$ of those $v \in [0,u]$ such that $0$ is dentable in $[0,v]$ is upward directed (and nonempty because of (i)).
(iii)
Show that the supremum of $A$ must be $u$, and that it must belong to $A$.

To prove (i), first we need to show that, for every $\varepsilon >0$, there exists $v \in [0,u]$ such that $\|v\|\ge \|u\|-\varepsilon$ and $0\not\in \co([0,v]\setminus B(0,\varepsilon ))$. Note that this is very similar to saying that $0$ is dentable in some interval, but letting the interval vary with $\varepsilon$; later we shall take the biggest of these intervals.

Suppose this were not true for some $\varepsilon >0$. Then we could construct a decreasing sequence $(v_n)$ with $v_n \in [0,u]$, $\|v_n\|\ge \|u\|-\varepsilon (1-2^{-n})$ and $\|v_n - v_{n+1}\|\ge \varepsilon$. Indeed, set $v_0=u$, and suppose that $v_n$ has already been constructed. Then, by assumption, $0\in \co([0,v_n]\setminus B(0,\varepsilon ))$, so there exist finitely many $x_i \in [0,v_n]$, with $\|x_i\|\ge \varepsilon$, and positive scalars $\lambda _i$ with $\sum_{i}^{}\lambda _i=1$, such that $\big\|\sum_{i}^{}\lambda _ix_i\big\|\le \varepsilon 2^{-n-1}$. Thus, since $v_n\ge \sum_{i}^{}\lambda _i x_i$, $\big\|v_n - \sum_{i}^{}\lambda _ix_i\big\|\ge \|v_n\|-\varepsilon 2^{-n-1}$, which readily implies that there must exist at least one $x_i$ for which $\|v_n - x_i\|\ge \|v_n\|-\varepsilon 2^{-n-1}$. Setting $v_{n+1}=v_n-x_i$ completes the construction, since $v_{n+1}\le v_n$, \[\|v_{n+1}\|\ge \|v_n\|-\varepsilon 2^{-n-1}\ge \|u\|-\varepsilon (1-2^{-n}+2^{-n-1})=\|u\|-\varepsilon (1-2^{-n-1}),\] and $\|v_n - v_{n+1}\|=\|x_i\|\ge \varepsilon$. But then the sequence $(v_n)$ is not convergent, contradicting the fact that $[0,u]$ is weakly compact and Theorem 4.4.

We have thus proved that for every $\varepsilon >0$, there exists $v \in [0,u]$ such that $\|v\|\ge \|u\|-\varepsilon$ and $0\not\in \co([0,v]\setminus B(0,\varepsilon ))$. This makes it possible to construct a decreasing sequence $(v_n)$ in $[0,u]$ such that $\|v_n\|\ge \|u\|-\frac{1}{2}(1-2^{-n})\|u\|$ and $0 \not\in \co([0,v_n]\setminus B(0,2^{-n}))$ for $n \geq 1$. Indeed, start with $v_0=u$ and, if $v_n$ has been constructed, apply the fact that we have proved to the interval $[0,v_{n}]$ (which is weakly compact, being a weakly closed subset of $[0,u]$) with $\varepsilon =\min\{2^{-n-1},2^{-n-2}\|u\|\}$: there exists $v_{n+1}\in [0,v_n]$ with $0 \not\in \co([0,v_{n+1}]\setminus B(0,2^{-n-1}))$ and such that \[\|v_{n+1}\|\ge \|v_n\|-\varepsilon \ge \|u\|-\frac{1}{2}(1-2^{-n})\|u\|-\varepsilon \ge \|u\|-\frac{1}{2}(1-2^{-n-1})\|u\|.\] If $v=\lim_n v_n$, which exists by Theorem 4.4, and in fact is the infimum of $(v_n)$, then $\|v\|\ge \frac{1}{2}\|u\|$ and $0$ is dentable in $[0,v]$. This completes step (i) of the proof.

To proceed to step (ii), let $A$ be the set of those $v \in [0,u]$ such that $0$ is dentable in $[0,v]$. In (i), we proved that this set is nonempty. Let us show that it is upward directed for the order inherited from $E$. Suppose that $v_1,v_2 \in [0,u]$ are such that $0$ is dentable both in $[0,v_1]$ and $[0,v_2]$. We first show that $0$ is also dentable in $[0,v_1+v_2]$. Let $\varepsilon > 0$. Recall that, by the Riesz decomposition theorem, every $x \in [0,v_1+v_2]$ can be decomposed as $x=x_1+x_2$ with $x_1 \in [0,v_1]$ and $x_2 \in [0,v_2]$. Moreover, $\|x\|\le 2\max(\|x_1\|,\|x_2\|)$. It follows that \[\begin{align*} [0, v_1+v_2]\setminus B(0,2\varepsilon )&\subseteq ([0,v_1]\setminus B(0,\varepsilon )+[0,v_2]) \\ &\qquad \cup ([0,v_1]+[0,v_2]\setminus B(0,\varepsilon ))\\ &\subseteq \co(\co([0,v_1]\setminus B(0,\varepsilon ))+[0,v_2]) \\ &\qquad \cup ([0,v_1]+\co([0,v_2]\setminus B(0,\varepsilon ))). \end{align*}\] Since $0\not\in \co([0,v_i]\setminus B(0,\varepsilon ))$, for $i=1,2$, the previous chain of inclusions implies that $0$ is a dentable point of $[0,v_1+v_2]$. Since $0$ is then also dentable in $[0,v']$ for every $v' \in [0,v_1+v_2]$, it is dentable in $[0,v_1 \vee v_2]$, and $v_1 \vee v_2 \in [0,u]$ is an upper bound of $v_1$ and $v_2$. Hence $A$ is upward directed.

Therefore, we can think of $A$ as an increasing net. By Theorem 4.4, it converges to its supremum $w$. We have thus reached step (iii) of the proof. Let us show that $w \in A$, that is, that $0$ is a dentable point of $[0,w]$. Let $\varepsilon >0$ be small enough (otherwise the argument is vacuous). Since $w$ is also the limit of $A$, there exists $v \in A$ such that $\|w-v\|<\varepsilon$. Now let $x_i \in [0,w]$ be finitely many elements with $\|x_i\|\ge 2\varepsilon$, and let $\alpha _i$ be positive numbers with $\sum_{i}^{}\alpha _i=1$. Note that \[\|x_i \wedge v\|\ge \|x_i\|- \|x_i-x_i \wedge v\|\ge 2\varepsilon -\|w-v\|\ge \varepsilon,\] where we are using that \[0 \leq x_i - x_i \wedge v = (x_i - v)_+ \leq (w - v)_+ = w - v,\] because $x_i \leq w$, and this implies the corresponding inequality for the norms. Since $0$ is a dentable point of $[0,v]$, it follows that \[\Big\|\sum_{i}^{}\alpha _ix_i\Big\|\ge \Big\|\sum_{i}^{}\alpha _i x_i\wedge v\Big\|\ge \delta\] for some $\delta >0$. Thus $0$ is also a dentable point of $[0,w]$, that is, $w \in A$. Finally, if $w<u$, then there would exist some nonzero $t \in [0,u-w]$ such that $0$ is dentable in $[0,t]$, by part (i) of the proof. But then, by what we proved in step (ii), $0$ is dentable in $[0,w+t]$, and $w + t \leq u$, so $w+t \in A$, contradicting the fact that $w$ is the supremum.□

Lemma 5.3. Let $E$ be a Banach lattice, and let $u \in E_+$ be such that $[0,u]$ is weakly compact. For $n \in \N$, let $H_n=\{\, x \in E_+ : \|x\wedge u\|\ge n^{-1} \, \}$. Then for every $n$, there exists $m \in \N$ such that $\co(H_n)\subseteq H_m$.

Proof. If $x_i \in H_n$ are finitely many elements, and $\alpha _i$ are positive scalars with $\sum_{i}^{}\alpha _i=1$, then \[\sum_{i}^{}\alpha _i (x_i\wedge u) \in \co([0,u]\setminus B(0,1/n)).\] By Lemma 5.2, there exists $m \in \N$ such that $y \in \co([0,u]\setminus B(0,1/n))$ implies $\|y\|\ge m^{-1}$. Therefore \[\bigg\|\bigg(\sum_{i}^{}\alpha _ix_i\bigg) \wedge u\bigg\|\ge \Big\|\sum_{i}^{}\alpha _i(x_i\wedge u)\Big\|\ge m^{-1}.\] This shows that $\rawco(H_n)\subseteq H_m$, and taking closures yields the desired result.□

Lemma 5.4. Let $E$ be an order continuous Banach lattice, and let $A,B\subseteq E_+$ be bounded closed convex sets. Then $\rawco(A\cup B)$ is closed, that is, $\co(A\cup B)=\rawco(A\cup B)$.

Proof. If $A$ or $B$ is empty there is nothing to prove, so assume that both are nonempty. Since $A$ and $B$ are convex, every element of $\rawco(A \cup B)$ is of the form $\lambda a + (1 - \lambda) b$ with $a \in A$, $b \in B$ and $\lambda \in [0,1]$; in other words, \[\rawco(A \cup B) = \bigcup_{\lambda \in [0,1]} \big(\lambda A + (1 - \lambda) B\big).\] Let $x \in \co(A \cup B)$. Then there exist $a_n \in A$, $b_n \in B$ and $\lambda_n \in [0,1]$ such that \[x = \lim_n \lambda_n a_n + (1 - \lambda_n) b_n.\] Passing to a subsequence, we may assume that $\lambda_n$ converges to some $\lambda \in [0,1]$. Since $A$ and $B$ are bounded, $\|(\lambda_n - \lambda) a_n\| \to 0$ and $\|(\lambda_n - \lambda) b_n\| \to 0$, so that \[x = \lim_n \big(\lambda a_n + (1 - \lambda) b_n\big).\] Let $f_n = \lambda a_n$ and $g_n = (1 - \lambda) b_n$, which are positive, and note that $x \geq 0$ because $E_+$ is closed. The idea is to show that $(f_n)$ is almost inside the interval $[0,x]$, and then use weak compactness of this interval. Indeed, since $g_n \geq 0$, \[0 \leq f_n - f_n \wedge x = (f_n - x)_+ \leq (f_n + g_n - x)_+ \leq |f_n + g_n - x|,\] and therefore \[\|f_n - f_n \wedge x\| \leq \|f_n + g_n - x\| \to 0.\] Since $E$ is order continuous, the interval $[0,x]$ is weakly compact by Theorem 4.3, so the sequence $(f_n \wedge x)$ has a weak cluster point $f \in [0,x]$; that is, there is a subnet $(f_{n_\beta} \wedge x)_\beta$ that converges weakly to $f$. By the previous estimate, the subnet $(f_{n_\beta})_\beta$ also converges weakly to $f$. The set $\lambda A$ is closed and convex, hence weakly closed by Mazur’s theorem, so $f \in \lambda A$. Consequently, \[g_{n_\beta} = (f_{n_\beta} + g_{n_\beta}) - f_{n_\beta} \to x - f \quad \text{weakly},\] and, by the same argument, $g = x - f \in (1 - \lambda) B$. Thus \[x = f + g \in \lambda A + (1 - \lambda) B \subseteq \rawco(A \cup B),\] which shows that $\rawco(A \cup B)$ is closed.□

The next lemma is where sets that are not order dentable enter the picture. It says that, if $A$ is not order dentable, then the part of $A$ that is far from $0$ in the weak sense of $H_n$ is negligible: it cannot help to recover $A$ from a smaller closed convex set by taking closed convex hulls. In the last two lemmas, $E$ denotes an order continuous Banach lattice with a quasi-interior point $u \in E_+$, and \[H_n = \{ x \in E_+ : \|x \wedge u\| \geq 1/n \},\] for $n \in \N$.

Lemma 5.5. Let $A \subseteq E_+$ be a nonempty closed bounded convex set which is not order dentable, let $n \in \N$, and let $B \subseteq A$ be a closed convex set such that \[A = \co\big((A \cap H_n) \cup B\big).\] Then $B = A$.

Proof. By Theorem 4.3, the interval $[0,u]$ is weakly compact, so by Lemma 5.3 there exists $m \in \N$ such that $\co(H_n) \subseteq H_m$. First, $B$ is nonempty. Otherwise $A = \co(A \cap H_n) \subseteq \co(H_n) \subseteq H_m$, so every slice of $A$ is contained in $H_m$, and $A$ would be order dentable.

Suppose, for a contradiction, that $B \neq A$, and let $a_0 \in A \setminus B$. By the Hahn–Banach separation theorem, there exists $f \in E^*$ such that $f(a_0) > \sup_{b \in B} f(b)$; rescaling, we may assume that $\|f\| = 1$. Then $M(A, f) > M(B, f)$, so we can choose $\alpha > 0$ such that \[4 \alpha < M(A, f) - M(B, f).\] We claim that the slice $S(A, f, \alpha)$ is contained in $H_{4m}$, which contradicts the fact that $A$ is not order dentable.

Let $x \in S(A, f, \alpha)$. By assumption, $x \in \co((A \cap H_n) \cup B)$, so there exists $y \in \rawco((A \cap H_n) \cup B)$ such that \[\|x - y\| \leq \min\{\alpha, (4m)^{-1}\}.\] Since $B$ is convex, we can write $y = \lambda a + (1 - \lambda) b$ with $\lambda \in [0,1]$, $a \in \rawco(A \cap H_n)$ and $b \in B$. On the one hand, since $\|f\| = 1$, \[f(y) \geq f(x) - \|x - y\| > M(A, f) - 2\alpha.\] On the other hand, $a \in A$ because $A$ is convex, so \[f(y) = \lambda f(a) + (1 - \lambda) f(b) \leq \lambda M(A, f) + (1 - \lambda) M(B, f).\] Combining both inequalities, \[(1 - \lambda) \big(M(A, f) - M(B, f)\big) < 2 \alpha < \frac{M(A, f) - M(B, f)}{2},\] which shows that $\lambda \geq 1/2$. Since $b \geq 0$, this gives $y \geq \lambda a \geq a/2 \geq 0$, and therefore \[y \wedge u \geq \frac{a}{2} \wedge u \geq \frac{a}{2} \wedge \frac{u}{2} = \frac{1}{2} (a \wedge u).\] Moreover, $a \in \co(H_n) \subseteq H_m$, so that \[\|y \wedge u\| \geq \frac{1}{2} \|a \wedge u\| \geq \frac{1}{2m}.\] Finally, using the standard inequality $|x \wedge u - y \wedge u| \leq |x - y|$, it follows that \[\|x \wedge u\| \geq \|y \wedge u\| - \|x - y\| \geq \frac{1}{2m} - \frac{1}{4m} = \frac{1}{4m},\] that is, $x \in H_{4m}$. This proves the claim and completes the proof.□

The last lemma is the building block of the construction in the main proof, and it is where all the previous lemmas come together. Start with a point $x$ of $A$ where some functional $f$ attains its maximum, and with a slice $S$ of $A$ determined by $f$. The lemma says that, up to an arbitrarily small error, $x$ can be split as a convex combination of points $y_i$ which are again points where some functional $f_i$ attains its maximum, and whose slices are contained in $S$ but avoid $H_n$.

Lemma 5.6. Let $A \subseteq E_+$ be a nonempty closed bounded convex set which is not order dentable. Let $x \in A$ and $f \in E^*$ be such that $f(x) = M(A, f)$, let $\alpha, \varepsilon > 0$ and $n \in \N$, and let $S = S(A, f, \alpha)$. Then there exist $p \in \N$, $y_1, \ldots, y_p \in A$, $f_1, \ldots, f_p \in E^*$, $\alpha_1, \ldots, \alpha_p > 0$ and $\lambda_1, \ldots, \lambda_p \geq 0$ such that, for every $i = 1, \ldots, p$,

(i)
$f_i(y_i) = M(A, f_i)$,
(ii)
$S(A, f_i, \alpha_i) \subseteq S$,
(iii)
$S(A, f_i, \alpha_i) \cap H_n = \emptyset$,

and moreover

(iv)
$\sum_{i=1}^p \lambda_i = 1$ and $\big\| x - \sum_{i=1}^p \lambda_i y_i \big\| < \varepsilon$.

Proof. Let $L = A \cap H_n$. If $(A \setminus S) \cup L$ is empty, then the lemma holds trivially, so we may assume that $(A \setminus S) \cup L$ is nonempty. Let $Z$ be the set of those $z \in A$ for which there exists $g \in E^*$ such that \[g(z) = M(A, g) > M\big((A \setminus S) \cup L, g\big).\] In words, $Z$ consists of the support points of $A$ whose maximum is strictly above its values on $A \setminus S$ and on $L$. The proof has the following steps:

(i)
Using the Bishop–Phelps theorem, show that $A = \co((A \setminus S) \cup L \cup Z)$.
(ii)
Using Lemma 5.5, get rid of $L$, that is, show that $A = \co((A \setminus S) \cup Z)$.
(iii)
Using Lemma 5.4, show that $x \in \co(Z)$, and deduce the result.

For step (i), let \[D = \co\big((A \setminus S) \cup L \cup Z\big),\] which is a nonempty closed convex subset of $A$, and suppose, for a contradiction, that $D \neq A$. By Corollary 3.7 there exist $g \in E^*$ and $z \in A$ such that $M(D,g) < M(A, g) = g(z)$. Since $(A \setminus S) \cup L \subseteq D$, this shows that $z \in Z \subseteq D$, and so $g(z) \leq M(D, g)$, a contradiction. Hence $D = A$.

For step (ii), put \[B = \co\big((A \setminus S) \cup Z\big),\] which is a closed convex subset of $A$. Then $A=\co(B\cup L)$ and, since $A$ is not order dentable, Lemma 5.5 shows that $B = A$.

For step (iii), note that every element $y$ of $\co(A \setminus S)$ satisfies $f(y) \leq M(A, f) - \alpha$, because this holds on $A \setminus S$ and the set $\{y \in E : f(y) \leq M(A, f) - \alpha\}$ is closed and convex. By Lemma 5.4 applied to the bounded closed convex sets $\co(A \setminus S)$ and $\co(Z)$, \[A = \co\big(\co(A \setminus S) \cup \co(Z)\big) = \rawco\big(\co(A \setminus S) \cup \co(Z)\big),\] so we can write $x = \lambda a + (1 - \lambda) z$ with $\lambda \in [0,1]$, $a \in \co(A \setminus S)$ and $z \in \co(Z)$. Then \[\begin{align*} M(A, f) = f(x) &= \lambda f(a) + (1 - \lambda) f(z) \\ &\leq \lambda \big(M(A, f) - \alpha\big) + (1 - \lambda) M(A, f) = M(A, f) - \lambda \alpha, \end{align*}\] which forces $\lambda = 0$, that is, $x \in \co(Z)$.

Finally, since $x \in \co(Z)$, there exist $p \in \N$, $y_1, \ldots, y_p \in Z$ and $\lambda_1, \ldots, \lambda_p \geq 0$ with $\sum_{i=1}^p \lambda_i = 1$ such that \[\Big\| x - \sum_{i=1}^p \lambda_i y_i \Big\| < \varepsilon,\] which is (iv). For each $i$, since $y_i \in Z$, there exists $f_i \in E^*$ such that \[f_i(y_i) = M(A, f_i) > M\big((A \setminus S) \cup L, f_i\big),\] which is (i), and we can choose $\alpha_i > 0$ such that \[M(A, f_i) - \alpha_i \geq M\big((A \setminus S) \cup L, f_i\big).\] Then the slice $S(A, f_i, \alpha_i)$ is disjoint from $(A \setminus S) \cup L$. Since it is contained in $A$, it is contained in $S$, which is (ii), and it is disjoint from $H_n$, because $L = A \cap H_n$, which is (iii).□

Proof 2. Suppose that $E$ is a Banach lattice with the Krein–Milman property. We want to show that $E$ has the Radon–Nikodým property. As already noted in Section 3, $E$ cannot contain $c_0$, and therefore by Theorem 4.6 it is a KB-space, hence order continuous. Thus, according to Theorem 4.3, its intervals are weakly compact, and therefore we can use all the machinery developed in the previous lemmas.

Moreover, by the order-theoretic characterization of the RNP (Theorem 4.8), since $E$ does not contain $c_0$, $E$ has the RNP if and only if it is order dentable. Suppose, in order to reach a contradiction, that $E$ is not order dentable. Then there exist a closed sublattice $F$ of $E$ with a quasi-interior point and a closed bounded convex set $A \subseteq F_+$ with $A \neq \{0\}$ that is not order dentable in $F$. Since $F$ is a closed subspace of $E$, it also has the KMP, so everything said above applies to $F$ as well. Hence we may replace $E$ by $F$, and assume that $E$ has a quasi-interior point $u \in E_+$, which we fix from now on (the sets $H_n$ are defined in terms of this $u$), and that $A \subseteq E_+$ is a closed bounded convex set with $A \neq \{0\}$ which is not order dentable. We will use Lemma 5.6 to construct an increasing family of finite algebras $\Sigma _n$ of $[0,1]$, and $\Sigma _n$-measurable functions $X_n\colon [0,1]\to A$, $f_n\colon [0,1]\to E^{*}$, and $\alpha _n\colon [0,1]\to \R_+$, such that, for every $t \in [0,1]$ and $n \in \N$,

(a)
$f_n(t)\big(X_n(t)\big) = M\big(A, f_n(t)\big)$,
(b)
$S\big(A, f_n(t), \alpha_n(t)\big) \subseteq S\big(A, f_{n-1}(t), \alpha_{n-1}(t)/2\big)$,
(c)
$S\big(A, f_n(t), \alpha_n(t)\big) \cap H_n = \emptyset$,
(d)
$\big\| X_{n-1}(t) - \mathbb{E}_{n-1}(X_n)(t) \big\| < 2^{-n}$, where $\mathbb{E}_n = \mathbb{E}(\,\cdot \mid \Sigma_n)$ denotes the conditional expectation with respect to $\Sigma_n$.

These functions will keep track of successive applications of Lemma 5.6. From (a), it will only be relevant to us that $X_n(t) \in S\big(A, f_n(t), \alpha_n(t)\big)$. From (b), what is important is that $\overline{S\big(A, f_n(t), \alpha_n(t)\big)} \subseteq S\big(A, f_{n-1}(t), \alpha_{n-1}(t)\big)$; the $\alpha_{n-1}(t)/2$ is only chosen so that this inclusion holds.

Let us explain the construction in detail. First, the base step $n=0$. By Bishop–Phelps, there exists a functional $f_0 \in E^{*}$ supporting $A$ at $x_0$ with $M(A, f_0) > 0$. Indeed, since $A \neq \{0\}$, there exist $a \in A$ and $h \in E^*$ with $h(a) > 0$; if $r = \sup_{y \in A} \|y\|$ and $f_0$ is a support functional of $A$ with $\|f_0 - h\| < h(a)/(2r)$, then $M(A, f_0) \geq f_0(a) > h(a)/2 > 0$. Put $\Sigma _0=\{\emptyset,[0,1]\}$, $X_0(t)=x_0$, $f_0(t)=f_0$, and $\alpha _0(t)=\alpha _0$, where $\alpha_0 = M(A, f_0)$. Then only the condition in (a) makes sense, and it is obviously satisfied. The requirement $M(A, f_0) > 0$ will be used at the end of the proof to guarantee that the set we construct is not $\{0\}$.

Since most conditions are vacuous in this base step $n=0$, let us also do the case $n=1$ explicitly. Start by applying Lemma 5.6 to $A$, $x_0 \in A$, $f_0 \in E^{*}$, $\alpha =\alpha _0/2$ (so that $S=S(A,f_0,\alpha _0/2)$), $\varepsilon =2^{-1}$ and $n=1$. The lemma yields $p \in \N$, $y_1,\ldots ,y_p \in A$, $f_1,\ldots ,f_p \in E^{*}$, $\alpha _1,\ldots ,\alpha _p >0$, and $\lambda _1,\ldots ,\lambda _p\ge 0$ with $\sum_{i=1}^{p}\lambda _i=1$, satisfying conditions (i) to (iv) of the lemma. Discarding the indices $i$ with $\lambda_i = 0$, which do not affect (iv), we may assume that $\lambda_i > 0$ for every $i$. Define $\Sigma _1$ to be the $\sigma$-algebra of $[0,1]$ generated by $p$ consecutive intervals $I_1,\ldots ,I_p$ partitioning $[0,1]$, of lengths $\lambda _1,\ldots ,\lambda _p$, respectively. Define also \[X_1=\sum_{i=1}^{p}y_i \chi _{I_i}, \quad f_1 = \sum_{i=1}^{p} f_i \chi_{I_i}, \quad \alpha_1 = \sum_{i=1}^{p} \alpha _i \chi_{I_i}.\] These are, by definition, $\Sigma _1$-measurable. Moreover, if $t \in I_{i_0}$, then $X_1(t) = y_{i_0}$, $f_1(t) = f_{i_0}$ and $\alpha_1(t) = \alpha _{i_0}$, so that

(a)
$f_1(t)\big(X_1(t)\big) = M\big(A, f_1(t)\big)$ is condition (i) of the lemma;
(b)
$S\big(A, f_1(t), \alpha_1(t)\big) \subseteq S(A, f_0, \alpha_0/2) = S\big(A, f_0(t), \alpha_0(t)/2\big)$ is condition (ii);
(c)
$S\big(A, f_1(t), \alpha_1(t)\big) \cap H_1 = \emptyset$ is condition (iii);
(d)
since $\Sigma_0$ is trivial, $\mathbb{E}_0(X_1)$ is the constant function \[\mathbb{E}_0(X_1)(t) = \int_0^1 X_1(s) \, ds = \sum_{i=1}^{p} \lambda_i y_i,\] and so condition (iv) gives \[\big\| X_0(t) - \mathbb{E}_0(X_1)(t) \big\| = \Big\| x_0 - \sum_{i=1}^{p} \lambda_i y_i \Big\| < 2^{-1}.\]

Now for the general induction step. Suppose $\Sigma _n, X_n, f_n, \alpha _n$ have already been constructed, for some $n\ge 1$. Let $I_1,\ldots ,I_k$ be the atoms of the finite $\sigma$-algebra $\Sigma _n$ (which, as we will see in the construction, are actually intervals). Let $x_j$, $f_j$ and $\alpha_j$ denote the constant values of $X_n$, $f_n$ and $\alpha_n$ on $I_j$. Apply Lemma 5.6 to $A$, $x_j$, $f_j$, $\alpha =\alpha _j/2$ (so that $S=S(A,f_j,\alpha _j/2)$), $\varepsilon =2^{-n-1}$ and $n+1$ in place of $n$. This yields $p_j$ points $y_{j 1},\ldots ,y_{j p_j}\in A$, functionals $f_{j1}, \ldots, f_{jp_j} \in E^*$, and numbers $\alpha _{j1}, \ldots, \alpha _{jp_j} > 0$ and $\lambda_{j1}, \ldots, \lambda_{jp_j} \geq 0$ with $\sum_{i=1}^{p_j} \lambda_{ji} = 1$, satisfying conditions (i) to (iv) of the lemma; this is possible because $f_j(x_j) = M(A, f_j)$ by (a). As before, we may assume that $\lambda_{ji} > 0$ for every $i$. Divide $I_j$ into $p_j$ consecutive intervals $I_{j1}, \ldots, I_{jp_j}$ of lengths $\lambda_{j1} |I_j|, \ldots, \lambda_{jp_j} |I_j|$, respectively, where $|I|$ denotes the length of an interval $I$. Define $\Sigma_{n+1}$ to be the $\sigma$-algebra of $[0,1]$ generated by the intervals $I_{ji}$, for $j = 1, \ldots, k$ and $i = 1, \ldots, p_j$, which refines $\Sigma_n$, and define \[X_{n+1} = \sum_{j=1}^{k} \sum_{i=1}^{p_j} y_{ji} \chi_{I_{ji}}, \quad f_{n+1} = \sum_{j=1}^{k} \sum_{i=1}^{p_j} f_{ji} \chi_{I_{ji}}, \quad \alpha_{n+1} = \sum_{j=1}^{k} \sum_{i=1}^{p_j} \alpha_{ji} \chi_{I_{ji}}.\] These are $\Sigma_{n+1}$-measurable, and for every $t \in I_{ji}$ conditions (a), (b) and (c) for $n + 1$ are conditions (i), (ii) and (iii) of the lemma, respectively, because $f_n(t) = f_j$ and $\alpha_n(t) = \alpha _j$. As for (d), the atom of $\Sigma_n$ that contains $t$ is $I_j$, so that \[\mathbb{E}_n(X_{n+1})(t) = \frac{1}{|I_j|} \int_{I_j} X_{n+1}(s) \, ds = \sum_{i=1}^{p_j} \lambda_{ji} y_{ji},\] and condition (iv) gives \[\big\| X_n(t) - \mathbb{E}_n(X_{n+1})(t) \big\| = \Big\| x_j - \sum_{i=1}^{p_j} \lambda_{ji} y_{ji} \Big\| < 2^{-n-1}. \tag{1}\] This completes the construction.

According to (d), $X_n$ is not really a martingale with respect to the filtration $(\Sigma _n)$… but we can fix that by passing to an appropriate limit. But why do we even want to get a martingale? Because, as (1) shows, the conditional expectations are a way to keep track of convex combinations. Right now, (d) guarantees that $X_n(t)$ is almost a convex combination of $X_{n+1}(s)$, for appropriate values of $s$. Passing to the limit to get a martingale will remove this approximation and turn it into an exact equality.

It is time to work out the details. Since $A$ is bounded, all the functions $X_n$ belong to $L_\infty([0,1]; E)$, so by Corollary 2.6 and (d), \[\begin{aligned} \big\| \mathbb{E}_n(X_{k+1}) - \mathbb{E}_n(X_k) \big\|_\infty &=\big\| \mathbb{E}_n(\mathbb{E}_k(X_{k+1}) - X_k) \big\|_\infty \\ &\leq \big\| \mathbb{E}_k(X_{k+1}) - X_k \big\|_\infty \leq 2^{-k-1}, \end{aligned}\] for $k\ge n$. Therefore, for $l > k \geq n$, \[\big\| \mathbb{E}_n(X_l) - \mathbb{E}_n(X_k) \big\|_\infty \leq \sum_{m=k}^{l-1} 2^{-m-1} < 2^{-k}.\] Hence $(\mathbb{E}_n(X_k))_{k\ge n}$ is uniformly Cauchy in $L_\infty ([0,1],\Sigma _n;E)$. Let $Y_n$ be its limit. According to Corollary 2.6, $\mathbb{E}_n$ is continuous, so \[\mathbb{E}_n(Y_{n+1}) = \lim_k \mathbb{E}_n\big(\mathbb{E}_{n+1}(X_k)\big) = \lim_k \mathbb{E}_n(X_k) = Y_n,\] and $(Y_n)$ is indeed a martingale with respect to the filtration $(\Sigma _n)$. Moreover, by construction \[X_k(s)\in S(A,f_n(s),\alpha _n(s)/2)\] for all $k\ge n$, and if $t$ belongs to the same atom in $\Sigma _n$ as $s$, then $f_n(s)=f_n(t)$ and $\alpha _n(s)=\alpha _n(t)$. Thus, being $\mathbb{E}_n(X_k)(t)$ a convex combination of $X_k(s)$, for $s$ in the same atom, with respect to $\Sigma _n$, as $t$, it follows that \[\mathbb{E}_n(X_k)(t) \in S(A,f_n(t),\alpha _n(t)/2).\] By taking the limit in $k$, it follows that $Y_n(t)$ lies in the closure of $S(A,f_n(t),\alpha _n(t)/2)$, and therefore that $Y_n(t)\in S(A,f_n(t), \alpha _n(t))$.

Consider now the closed convex hull of all the values of the martingale, \[C = \co\big(\{ Y_n(t) : n \in \N,\ t \in [0,1] \}\big).\] Since $Y_n(t) \in S(A, f_n(t), \alpha_n(t)) \subseteq A$ and $A$ is closed and convex, $C$ is a nonempty closed bounded convex subset of $A$, and in particular $C \subseteq E_+$. Moreover, $C \neq \{0\}$: for every $t \in [0,1]$, $Y_0(t)$ lies in the closure of $S(A, f_0, \alpha_0/2)$, so that \[f_0\big(Y_0(t)\big) \geq M(A, f_0) - \frac{\alpha_0}{2} = \frac{M(A, f_0)}{2} > 0,\] because $\alpha_0 = M(A, f_0) > 0$; in particular, $Y_0(t) \neq 0$. We will see that $C$ has only $0$ as an extreme point and, therefore, it contradicts the Krein–Milman property.

But before that, let us try to make sense of this choice of $C$. The way to show that a point is not extreme is to write it as a convex combination of other points of the set. The martingale does exactly this, in a systematic way. Indeed, let $t \in [0,1]$, let $I$ be the atom of $\Sigma_n$ that contains $t$, and let $J_1, \ldots, J_q$ be the atoms of $\Sigma_{n+1}$ contained in $I$. Since $Y_{n+1}$ is constant on each $J_l$, say equal to $z_l$, the martingale property gives \[Y_n(t) = \mathbb{E}_n(Y_{n+1})(t) = \frac{1}{|I|} \int_I Y_{n+1}(s) \, ds = \sum_{l=1}^{q} \frac{|J_l|}{|I|} z_l,\] so every value of $Y_n$ is a convex combination of values of $Y_{n+1}$, which in turn are convex combinations of values of $Y_{n+2}$, and so on. In other words, the values of the martingale form a tree in $A$ in which every node is a convex combination of its successors, and $C$ is the smallest closed convex set containing this tree. This shows why passing to the limit to get a proper martingale was critical. Moreover, by (b) and (c), all the values of $Y_m$, for $m \geq n$, on an atom of $\Sigma_n$ lie in a single slice that is disjoint from $H_n$; that is, far down the tree, the points are close to $0$ in the order-theoretic sense measured by the sets $H_n$.

levels

Figure 3. The tree of values of a martingale $(Y_n)$ with respect to a filtration $(\Sigma_n)$. On the left, the atoms of $\Sigma_0, \Sigma_1, \ldots$ split $[0,1]$ into smaller and smaller intervals; on the right, $Y_n$ takes one value in $A$ on each atom of $\Sigma_n$. Hover over an atom $I$, or over a point: the value of $Y_n$ on $I$ is the convex combination of the values on the atoms $J_l \subseteq I$ of $\Sigma_{n+1}$, with weights $|J_l|/|I|$. The dashed region is the closed convex hull $C$ of all the values. Use the slider to add levels, and new for another martingale.

Let us now work out the details. Let $x \in C$ be nonzero, and let $u \in E_+$ be the quasi-interior point fixed above. Since $x\wedge n u \le n(x\wedge u)$, and $u$ is a quasi-interior point, it must be $0< x\wedge u$. Let $n \in \N$ be such that $x \in H_n$. Let $I_1,\ldots ,I_p$ be the atoms of $\Sigma _n$, and for $1\le l\le p$ let $C_l=\co ( \{\, Y_m(t) : m\ge n,t \in I_l \, \})$. By (b), \[Y_m(s)\in \overline{S(A,f_m(s),\alpha _m(s)/2)}\subseteq \overline{S(A,f_n(s),\alpha _n(s)/2)}=\overline{S(A,f_n(t),\alpha _n(t)/2)}\] for every $s,t \in I_l$ and $m\ge n$. This last set is a closed convex set contained in $S(A,f_n(t),\alpha _n(t))$. Hence $C_l\subseteq S(A,f_n(t),\alpha_n(t))$ for $1\le l\le p$; in particular, by (c), $C_l\cap H_n=\emptyset$, and $x \not\in C_l$.

We want to show that $C=\rawco\left( \bigcup_{l\le p} C_l \right)$. This will follow from a successive application of Lemma 5.4 once we show that \[C=\co\left( \bigcup_{l\le p} C_l \right).\] The non-trivial inclusion amounts to showing that $Y_m(t)$ is in the right-hand side of the equation for every $t \in [0,1]$ and every $m \in \N$. For $m\ge n$ this is obvious. For $m<n$, we have that $Y_m(t)=\mathbb{E}_m(Y_n)(t)$, which is a convex combination of the values of $Y_n$ at the atoms in $\Sigma _n$ that are descendants of the atom in $\Sigma _m$ that contains $t$. Hence $Y_m(t) \in \rawco\left( \bigcup_{l\le p} C_l \right)$. Thus $x \in C=\rawco\left( \bigcup_{l\le p} C_l \right)$ cannot be extreme, for if it were, it would belong to some $C_l$, which as we have already seen is not the case.

In conclusion, $C$ is a nonempty closed bounded convex subset of $E_+$ that is not the closed convex hull of its extreme points. Using it, we can even construct a nonempty closed bounded convex subset of $E_+$ that does not have extreme points. Indeed, use Bishop–Phelps to obtain a support functional $f \in E^{*}$ of $C$ with $M(C,f)>0$ (as in the base step of the construction); then $D=\{\, x \in C : f(x)=M(C,f) \, \}$ is extremal in $C$, yet it does not contain $0$, so it does not have any extreme point.□

AI disclosure and image credits

Claude Opus 5.5 was used to autocomplete text, to proofread, and to create the web version of the paper (including the embedded applications). The idea, exposition and most of the text are due to the author.

Portraits on the title page and in the margins, all cropped and converted to grayscale: J. Radon (public domain); O. Nikodym and S. Banach, monument in Kraków by Stefan Dousa, photo by Uuuxxxyyyzzz (CC BY-SA 4.0); O. Nikodym, the same monument, photo by Aw58 (CC BY 3.0); S. Bochner, about 1970, Woodson Research Center, Rice University (CC BY 3.0); P. Lévy and J. L. Doob, photos by Konrad Jacobs, Archives of the Mathematisches Forschungsinstitut Oberwolfach (CC BY-SA 2.0 DE); J. Lindenstrauss, photo by George Bergman (CC BY-SA 4.0); R. Phelps, photo by Konrad Jacobs, Archives of the Mathematisches Forschungsinstitut Oberwolfach (CC BY-SA 2.0 DE); F. Riesz (public domain); L. Kantorovich, photo by Andrei Bogdanov (CC BY 3.0); S. Banach (public domain); M. Krein, memorial plaque, photo by Sergey Marchenko (CC BY-SA 3.0); J. Bourgain, photo by George Bergman (CC BY-SA 4.0); M. Talagrand, photo by George Bergman, Archives of the Mathematisches Forschungsinstitut Oberwolfach (GFDL 1.2). The adapted images are shared under the same licenses.

References

[1] F. Albiac and N. J. Kalton. Topics in Banach space theory. Second. Vol. 233. Graduate Texts in Mathematics. With a foreword by Gilles Godefory. Springer, [Cham], 2016, pp. xx+508. MR: 3526021.
[2] C. D. Aliprantis and O. Burkinshaw. Positive operators. Reprint of the 1985 original. Springer, Dordrecht, 2006, pp. xx+376. MR: 2262133.
[3] E. Bishop and R. R. Phelps. “The support functionals of a convex set”. Proc. Sympos. Pure Math., Vol. VII. Amer. Math. Soc., Providence, RI, 1963, pp. 27–35. MR: 154092.
[4] S. Bochner. “Integration von Funktionen, deren Werte die Elemente eines Vektorraumes sind”. German. Fundam. Math. 20 (1933), pp. 262–276.
[5] J. Bourgain. “A geometric characterization of the Radon-Nikodým property in Banach spaces”. Compositio Math. 36.1 (1978), pp. 3–6. MR: 515034.
[6] J. Bourgain and M. Talagrand. “Dans un espace de Banach reticulé solide, la propriété de Radon-Nikodým et celle de Kreı̆n-Mil′man sont équivalentes”. Proc. Amer. Math. Soc. 81.1 (1981), pp. 93–96. MR: 589144.
[7] S. D. Chatterji. “Martingale convergence and the Radon-Nikodym theorem in Banach spaces”. Math. Scand. 22 (1968), pp. 21–41. MR: 246341.
[8] J. Diestel. Geometry of Banach spaces—selected topics. Lecture Notes in Mathematics, Vol. 485. Springer-Verlag, Berlin-New York, 1975, pp. xi+282. MR: 461094.
[9] J. L. Doob. Stochastic processes. John Wiley & Sons, Inc., New York; Chapman & Hall, Ltd., London, 1953.
[10] N. Ghoussoub and M. Talagrand. “Order dentability and the Radon-Nikodým property in Banach lattices”. Math. Ann. 243.3 (1979), pp. 217–225. MR: 548802.
[11] N. Ghoussoub. “Riesz-space-valued measures and processes”. Bull. Soc. Math. France 110.3 (1982), pp. 233–257. MR: 688033.
[12] R. E. Huff and P. D. Morris. “Dual spaces with the Krein-Milman property have the Radon-Nikodým property”. Proc. Amer. Math. Soc. 49 (1975), pp. 104–108. MR: 361775.
[13] M. Krein and D. Milman. “On extreme points of regular convex sets”. Studia Math. 9 (1940), pp. 133–138. MR: 4990.
[14] J. Lindenstrauss. “On extreme points in $l_{1}$”. Israel J. Math. 4 (1966), pp. 59–61. MR: 200701.
[15] J. Lindenstrauss and L. Tzafriri. Classical Banach spaces. II. Vol. 97. Ergebnisse der Mathematik und ihrer Grenzgebiete [Results in Mathematics and Related Areas]. Function spaces. Springer-Verlag, Berlin-New York, 1979, pp. x+243. MR: 540367.
[16] P. Meyer-Nieberg. Banach lattices. Universitext. Springer-Verlag, Berlin, 1991, pp. xvi+395. MR: 1128093.
[17] O. Nikodym. “Sur une généralisation des intégrales de M. J. Radon.” French. Fundam. Math. 15 (1930), pp. 131–179.
[18] J. Radon. “Theorie und Anwendungen der absolut additiven Mengenfunktionen.” German. Wien. Ber. 122 (1913), pp. 1295–1438.
[19] W. Schachermayer. “For a Banach space isomorphic to its square the Radon-Nikodým property and the Kreı̆n-Milman property are equivalent”. Studia Math. 81.3 (1985), pp. 329–339. MR: 808576.
[20] H. H. Schaefer. Banach lattices and positive operators. Die Grundlehren der mathematischen Wissenschaften, Band 215. Springer-Verlag, New York-Heidelberg, 1974, pp. xi+376. MR: 423039.
[21] J. Ville. Étude critique de la notion de collectif. Vol. 3. Monographies des Probabilités. Gauthier-Villars, Paris, 1939.