Banach Valued Integration and the Radon Nikodym Property — Examples

Form the finite simple approximants. For $N\ge 1$, put $s_N={\sum }_{n=1}^N{x}_n{\mathbf{1}}_{A_n}$. If $x_n\mathrm{\ne }0$, the finiteness of the displayed series forces $\mu (A_n)<\mathrm{\infty }$; zero levels need no finite-measure hypothesis. Thus every $s_N$ is an integrable simple function in the precise sense of [L1]. Pairwise disjointness gives $s_N(t)\to f(t)$ for every $t$: at most one summand is nonzero at any point.

Calculate the scalar approximation error. Pointwise disjointness gives $\parallel f-s_N\parallel ={\sum }_{n>N}\parallel x_n\parallel {\mathbf{1}}_{A_n}$. Applying monotone convergence in [L3] to its finite partial sums yields

Establish Bochner integrability and identify the unrestricted integral. The everywhere simple convergence in step 1.1 proves strong measurability, and step 2.1 gives integrability of the norm. Hence [L2] makes $f$ Bochner integrable. Moreover, $(s_N)$ is a defining $L^1$-simple approximation, so

Calculate every restricted integral and audit the boundary cases. [L1, L2, step 2.1, step 3.1] For measurable $E$, the functions ${\mathbf{1}}_E{s}_N$ approximate ${\mathbf{1}}_{E}f$ in $L^1$, since their error integral is at most the tail in step 2.1. The simple calculation from [L1] therefore gives ${\int }_{E}f={\sum }_n\mu (E\cap A_n)x_n$. If $E=\mathrm{\varnothing }$, if every $A_n$ is empty, or if every $x_n=0$, both sides are zero. A single nonzero level reduces to the defining simple-function formula. Infinite-measure zero levels cause no undefined product, while a nonzero level automatically has finite measure.

Verify that the target is a Banach space. Finite-dimensional Cauchy--Schwarz gives the triangle inequality after taking the supremum over finite $F$; homogeneity and definiteness are immediate, so the displayed formula is a norm. If $(y_n)$ is Cauchy in this norm, then every coordinate sequence $(y_n(t))$ is Cauchy. Let $y(t)={\lim }_n{y}_n(t)$. Given $\epsilon >0$, choose $N$ such that $\parallel y_m-y_N\parallel <\epsilon$ for $m\ge N$. For every finite $F$, passage to the scalar limit gives ${\sum }_{t\in F}\mathrm{\mid }y(t)-y_N(t){\mathrm{\mid }}^2\le {\epsilon }^2$. Taking the supremum shows $y-y_N\in H$ with norm at most $\epsilon$. Hence $y\in H$ and $y_n\to y$, so [L3] makes $H$ Banach. This also proves directly that $\parallel e_s-e_t\parallel =\sqrt{2}$ whenever $s\mathrm{\ne }t$.

Every scalar evaluation of the range has countable support. Fix $\phi \in H^{\ast }$ and put $a_t=\phi (e_t)$. For a finite $F\subseteq I$, apply $\phi$ to ${\sum }_{t\in F}\overline{a_t}{e}_t$ (with conjugation trivial over $\mathbb{R}$) to obtain

Prove weak measurability. The trace measure space on $I$ is complete: any subset of a trace-null set is an ambient subset of a Lebesgue-null set and hence is Lebesgue measurable by [L1]. For the fixed $\phi$, the scalar function $\phi \circ f:t\mapsto a_t$ vanishes off the countable null support from step 2.1. The inverse image of an open scalar set is either a subset of that support or the complement of such a subset, according as the open set omits or contains zero. Completeness makes every such inverse image measurable. Since $\phi$ was arbitrary, $f$ is weakly measurable.

Rule out an essentially separable range. Suppose there were a null $N\subseteq I$ and a separable closed subspace $Y\subseteq H$ containing every $e_t$ for $t\in I\setminus N$. The set $I\setminus N$ is uncountable: if it were countable, [L1] would make both it and $N$ null, contrary to $\mu (I)=1$. Let $D$ be an at most countable dense subset of $Y$. For each $t\in I\setminus N$, assign the first member of a fixed enumeration of $D$ lying within $\sqrt{2}\mathrm{/}3$ of $e_t$. The assignment is injective, because one point of $D$ cannot lie within that radius of two vectors at distance $\sqrt{2}$. This would make $I\setminus N$ countable, a contradiction. Thus the range is not essentially separably valued.

Conclude failure of strong measurability and audit boundaries. [A1, L2, step 1.1, step 3.1, step 4.1] The trace measure is complete, $H$ is Banach, and step 3.1 proves weak measurability, but step 4.1 disproves the other necessary condition in [L2]. Hence $f$ is not strongly measurable. Every coordinate vector has norm one; the zero functional has empty support and gives the constant zero scalar map; the real and complex cases are both covered by the finite coefficient calculation. The positive-measure interval, rather than a singleton or a null domain, is essential to the failed conclusion. AC is used through [L1], [L2], and the displayed simultaneous countability arguments only.

Obtain the vector-measure conclusions. Apply [L1] to $f$. It gives norm countable additivity of $E\mapsto {\nu }_f(E)$, absolute continuity with respect to $\mu$, and the equality $\mathrm{\mid }{\nu }_f\mathrm{\mid }(E)={\int }_E\parallel f\parallel$ for every measurable $E$. This is an equality of finite positive measures, not merely an upper estimate on $\parallel {\nu }_f(E)\parallel$.

Calculate the countably valued special case. Put $s_N={\sum }_{n\le N}{x}_n{\mathbf{1}}_{A_n}$. These are integrable simple functions and converge pointwise to $f$. Pairwise disjointness and monotone convergence in [L2] give $\int \parallel f-s_N\parallel ={\sum }_{n>N}\mu (A_n)\parallel x_n\parallel \to 0$, so [L2] makes $f$ Bochner integrable. Restricting the same approximation to $E$ and using the finite simple formula gives ${\nu }_f(E)={\sum }_n\mu (E\cap A_n)x_n$. Also $\parallel f\parallel ={\sum }_n\parallel x_n\parallel {\mathbf{1}}_{A_n}$ pointwise, so [L1] and the same scalar monotone-convergence calculation give $\mathrm{\mid }{\nu }_f\mathrm{\mid }(E)={\sum }_n\mu (E\cap A_n)\parallel x_n\parallel$.

Audit the examples at the boundaries. [L1, L2, step 1.1, step 2.1] For $E=\mathrm{\varnothing }$ both measures vanish. For $f=0$, the induced vector measure and its variation are both zero. With one nonzero level the two formulas read ${\nu }_f(E)=\mu (E\cap A)x$ and $\mathrm{\mid }{\nu }_f\mathrm{\mid }(E)=\mu (E\cap A)\parallel x\parallel$, exhibiting equality even when cancellation would make the norm of a multi-level vector sum smaller. A zero coefficient on an infinite-measure level contributes zero under the established simple-integral convention.

Propagate the choice assumption to the reflexivity supplier. By [A1], the assumed AC supplies the Countable Choice required by [L1]. No separate choice assumption is introduced.

Pass from Hilbert structure to RNP. A Hilbert space is complete for its inner-product norm, so [L1] makes $H$ reflexive. It is therefore a reflexive Banach space, and [L2], under the same AC hypothesis, gives the Radon--Nikodym property.

Audit the scope and degenerate cases. [A1, L1, L2, step 1.1, step 2.1] The argument applies to both real and complex scalar fields and to Hilbert spaces of arbitrary dimension and separability. For the zero Hilbert space, reflexivity and RNP are included in the two suppliers and the density condition is vacuous at zero vector measures. Completeness is essential to the word ``Hilbert'' here; no assertion is made for an incomplete inner-product space. The only choice propagation is AC to Countable Choice in step 1.1 and AC into [L2].

Fix an arbitrary slice with a strict margin. Let $S=S(B,\varphi ,\alpha )$ be a slice. By [L2], write $\varphi (x)={\sum }_n{a}_n{x}_n$. One has ${\sup }_{x\in B}\mathrm{Re}\varphi (x)=\parallel \varphi \parallel$: the upper bound is the dual-norm inequality, while multiplying any almost norming vector by a scalar of modulus one makes its value real and nonnegative. Choose $x\in S$ and set

Construct two points in the slice at distance two. The ${\mathrm{\ell }}^1$ truncation convergence in [L2] implies $a_k\to 0$, so take $k$ with $2\mathrm{\mid }{a}_k\mathrm{\mid }<\delta$. Retain all coordinates of $x$ except put $y_k=1$ and $z_k=-1$. A one-coordinate change preserves convergence to zero, and $\parallel y{\parallel }_{\mathrm{\infty }},\parallel z{\parallel }_{\mathrm{\infty }}\le 1$, so $y,z\in B$. Moreover,

Compute the diameter and deduce nondentability. The triangle inequality bounds the diameter of $B$, and thus of $S$, above by two. Step 2.1 attains two, so every slice has diameter exactly two. In particular no slice has diameter below one, and [L3] says that $B$ is not dentable. The set $B$ is nonempty, bounded, closed, and convex in the Banach space from [L1].

Audit zero, strict-boundary, and complex cases. [L2, L3, step 1.1, step 2.1, step 3.1] If $\varphi =0$, then the slice is all of $B$ and $e_k,-e_k$ are direct witnesses. For nonzero $\varphi$, the positive $\delta$ and strict inequality $2\mathrm{\mid }{a}_k\mathrm{\mid }<\delta$ keep both witnesses inside the slice rather than only on its boundary; a zero remote coefficient is harmless. In the complex case [L2] uses the bilinear pairing and [L3] uses $\mathrm{Re}\varphi$, so no conjugation is inserted. A singleton zero ball is not involved: $c_0$ contains every $e_k$.

Verify norm separability of the sequence space. Over $\mathbb{R}$, let $D$ be the finite-support sequences with rational coordinates; over $\mathbb{C}$, use coordinates in $\mathbb{Q}+i\mathbb{Q}$. For each support length the coordinate choices form a finite product of countable sets, and the union over all lengths is countable by [L1]. Given $a\in {\mathrm{\ell }}^1$ and $\epsilon >0$, first choose a finite truncation within $\epsilon \mathrm{/}2$ in ${\mathrm{\ell }}^1$, then approximate its finitely many coordinates so that the sum of coordinate errors is below $\epsilon \mathrm{/}2$. Thus $D$ is countable and dense, and ${\mathrm{\ell }}^1$ is norm separable.

Put the sequence-space side under the separable-dual theorem. By [L2], ${\mathrm{\ell }}^1$ is isometrically the continuous dual $c_0^{\ast }$ and is Banach. Step 1.1 supplies norm separability, so [L3], under the assumed AC, gives RNP to ${\mathrm{\ell }}^1$.

Contrast the nonatomic function space and audit scope. [A1, L4, step 2.1] The theorem [L4] gives the opposite conclusion for the real and complex Lebesgue quotient spaces $L^1([0,1],\lambda )$. This is not a contradiction: ${\mathrm{\ell }}^1$ consists of summable scalar sequences and is the separable dual $c_0^{\ast }$, whereas the second space is built over a nonatomic measure and has the explicit nondifferentiable indicator curve used in [L4]. The zero sequence and zero function occur in both spaces but do not determine a global geometric property. Both scalar fields are covered, and AC is propagated to [L3] and [L4], with Countable Choice used in the countability calculation of step 1.1.

Isolate what the scalar theorem supplies. For a real signed measure $\nu \ll \mu$ satisfying [L1]'s common finite-exhaustion hypotheses, [L1] supplies a scalar measurable $f$ with $\nu (E)={\int }_{E}fd\mu$ for every measurable $E$. When $\nu$ has finite variation, $f\in L^1$, and [L2] identifies $\mathrm{\mid }\nu \mathrm{\mid }(E)={\int }_E\mathrm{\mid }f\mathrm{\mid }d\mu$. Thus existence, uniqueness up to almost-everywhere equality, and scalar variation all live inside the ordered scalar theory.

Compare the vector quantifiers and density notion. For a Banach target $X$, [L3] begins with a norm-countably additive map $\nu :\mathcal{A}\to X$, not a signed scalar measure. Its bounded variation is the supremum of sums of vector norms over finite partitions. The requested density is an $X$-valued strongly measurable, norm-integrable Bochner function, and its integral must recover $\nu (E)$ for every $E$. None of these target-valued existence assertions follows merely by replacing absolute values with norms in step 1.1.

Locate the overlap without overclaiming. When $X=\mathbb{R}$, a norm-countably additive vector measure is a signed measure, bounded variation is finite scalar total variation, and scalar measurability/integrability is the real Bochner notion. On a finite control measure the constant exhaustion meets [L1], so the scalar theorem supplies this special RNP case, with [L2] supplying its variation formula. For a general $X$, [L3] remains a genuine extra geometric requirement.

Audit the boundaries and assumptions. [A1, L1, L3, step 3.1] The empty measurable space and zero scalar measure give zero densities in both settings. The zero Banach target has RNP trivially, but this says nothing about nonzero targets. The cited scalar theorem is real; no complex scalar theorem is silently extracted from it. Its sigma-finite-style common exhaustion is more general than the finite control measures in the RNP definition, while finite variation is what makes its scalar density $L^1$. AC is stated because [L1] requires it.

Fix the finite quotient-space model. Let ${\lambda }_I(E)=\lambda (E\cap I)$ on the ambient Lebesgue sigma-algebra. By [L1] this is a finite measure with ${\lambda }_I(I)=1$. We use the quotient $L^1({\lambda }_I)$ from [L2], so changes outside $I$ or on null endpoints do not change a class.

Prove uniform integrability of the dominated family. For $f\in K_g$, domination gives $\parallel f{\parallel }_1\le {\int }_{I}g$, uniformly in $f$. Given $\epsilon >0$, [L3] supplies $\delta >0$ such that ${\lambda }_I(E)<\delta$ implies ${\int }_{E}g<\epsilon$. Then ${\int }_E\mathrm{\mid }f\mathrm{\mid }\le {\int }_{E}g<\epsilon$ for every $f\in K_g$. Thus the two conditions in [L4] hold, so $K_g$ is uniformly integrable and relatively weakly compact.

Calculate the concentrating sequence. For every $n\ge 1$, the nonnegative simple-integral formula and interval length give

Fail uniform integrability and weak compactness. Taking, for example, $\epsilon =1\mathrm{/}2$, step 2.2 shows that no single $\delta >0$ works for the uniform absolute-continuity condition: choose $n>1\mathrm{/}\delta$. Therefore the family $\{f_n:n\ge 1\}$ is not uniformly integrable. The reverse implication in [L4] then shows that it is not relatively weakly compact.

Audit the endpoints and degenerate families. [A1, L1, L2, L3, L4, step 1.1, step 2.1, step 2.2, step 3.1] If $K_g=\mathrm{\varnothing }$, both conclusions in part 1 are vacuous. If $g=0$, every dominated $L^1$ class is zero, so the conclusion is the compact singleton case. Open, closed, or half-open spike intervals define the same class because their endpoint differences are null. The spikes are real and nonnegative; their obstruction is concentration on shrinking positive-measure sets, not unbounded $L^1$ norm. AC is used exactly through [L4] and to supply the Countable Choice in the Lebesgue model [L1].