On non-separable $L^1$-spaces of a vector measure

Abstract

Let $\kappa$ be an infinite cardinal. Let $\nu$ be a (countably additive Banach space-valued) vector measure defined on a $\sigma$-algebra $\Sigma$. We prove that if $\nu$ is homogeneous and $L^1(\nu )$ has density character $\kappa$, then there is a vector measure $\tilde{\nu }:\Sigma \rightarrow \ell ^\infty _{<}(\kappa )$ such that $L^1(\nu )=L^1(\tilde{\nu })$ with equal norms. Here $\ell ^\infty _{<}(\kappa )$ denotes the subspace of $\ell ^\infty (\kappa )$ consisting of all $(x_\alpha )_{\alpha <\kappa }\in \ell ^\infty (\kappa )$ such that $|\{\alpha<\kappa : |x_\alpha |>\varepsilon \}|<\kappa$ for every $\varepsilon >0$. In this way, we extend to the non-separable setting a result of Curbera corresponding to the case $\kappa =\omega$. Some other results on non-separable $L^1$ spaces of vector measures are given.

Appendix: Linear injections into \(L^1\) of a vector measure

As we mentioned in the introduction, given an uncountable set $\Gamma$, the space $\ell ^p(\Gamma )$ is not isomorphic to the $L^1$ space of a vector measure for any $p\ne 2$, see [26, Theorem 2.6]. However, we should note that any order continuous Banach lattice, like $\ell ^p(\Gamma )$, is lattice isometric to the $L^1$ space of a “vector measure” defined on a $\delta$ -ring (a structure which is weaker than $\sigma$-algebra), see [5, pp. 22–23].

On the other hand, for an uncountable set $\Gamma$, the space $\ell ^p(\Gamma )$ embeds isomorphically into the $L^1$ space of a finite measure if and only if $1<p\le 2$, see [10, Theorem 2.1]. For the range $2<p<\infty$ the situation is different:

Proposition 3.4

Let $\Gamma$ be a non-empty set and let Z be either $c_0(\Gamma )$ or $\ell ^p(\Gamma )$ for some $2<p<\infty$. If there is an injective operator from Z into $L^1(\nu )$ for some $\nu \in ca(\Sigma ,X)$, then $\Gamma$ is countable.

Proof

Let $S:Z \rightarrow L^1(\nu )$ be an injective operator. Let $\mu$ be a Rybakov control measure of $\nu$ and $i:L^1(\nu )\rightarrow L^1(\mu )$ the inclusion operator, which is injective. Then $T:=i\circ S: Z \rightarrow L^1(\mu )$ is injective as well. Since $Z^*=\ell ^q(\Gamma )$ for some $1\le q < 2$, the adjoint operator $T^*: L^\infty (\mu ) \rightarrow Z^*$ is compact, by a result of Rosenthal (see [27, p. 211, Remark 2]). By Schauder’s theorem, T is compact and so T(Z) is separable. Therefore, there is a countable set $\Delta \subseteq L^\infty (\mu )$ separating the points of T(Z). Since T is injective, the countable set $T^*(\Delta ) \subseteq Z^*$ separates the points of Z, hence $(Z^*,w^*)$ is separable. This clearly implies that $\Gamma$ is countable. $\square$

In particular, for any uncountable set $\Gamma$ the space $c_0(\Gamma )$ does not embed isomorphically into the $L^1$ space of a vector measure. This assertion can be extended to all infinite-dimensional C(K) spaces except $c_0$ itself, see Corollary 3.6 below.

A Banach space Z is said to be weakly countably determined (WCD) if there is a sequence $(K_n)$ of $w^*$-compact subsets of $Z^{**}$ such that, for every $z\in Z$ and $z^{**}\in Z^{**}{\setminus } Z$, there is $n\in \mathbb {N}$ such that $z\in K_n$ and $z^{**}\not \in K_n$. The class of WCD Banach spaces includes all weakly compactly generated spaces and their subspaces. For complete information on WCD spaces, we refer the reader to [11, Chapter 7].

A Banach space Z is said to have the Dunford–Pettis property if every weakly compact operator T from Z to a Banach space is Dunford–Pettis (i.e. T(C) is norm compact whenever $C \subseteq Z$ is weakly compact).

Proposition 3.5

Let Z be a WCD Banach space with the Dunford–Pettis property such that $Z^*$ contains no subspace isomorphic to $c_0$. If there is an injective operator from Z into $L^1(\nu )$ for some $\nu \in ca(\Sigma ,X)$, then Z is separable.

Proof

The proof is similar to that of Proposition 3.4. Fix an injective operator $S:Z \rightarrow L^1(\nu )$. Let $\mu$ be a Rybakov control measure of $\nu$, let $i:L^1(\nu )\rightarrow L^1(\mu )$ be the inclusion operator and define $T:=i\circ S: Z \rightarrow L^1(\mu )$. Observe that the adjoint $T^*: L^\infty (\mu ) \rightarrow Z^*$ is weakly compact, because $L^\infty (\mu )$ is a C(K) space and $Z^*$ contains no subspace isomorphic to $c_0$ (see e.g. [1, Theorem 5.5.3]). By Gantmacher’s theorem, T is weakly compact and so the Dunford–Pettis property of Z ensures that T is a Dunford–Pettis operator. Since every Dunford–Pettis operator from a WCD Banach space has separable range (see [28, Theorem 7.1]), T(Z) is separable. The rest of the proof follows the argument of Proposition 3.4, bearing in mind that a WCD Banach space is separable if (and only if) it has $w^*$-separable dual (see [28, Theorem 6.1] or [30, Corollary 2]). $\square$

For any compact Hausdorff topological space K, the Banach space C(K) has the Dunford–Pettis property (see e.g. [1, Theorem 5.4.5]) and its dual $C(K)^*$ contains no subspace isomorphic to $c_0$ (combine [1, Theorem 5.5.3] and [2, Theorem 4.68]). These facts and Proposition 3.5 yield the following:

Corollary 3.6

Let K be an infinite compact Hausdorff topological space. If C(K) is isomorphic to a subspace of $L^1(\nu )$ for some $\nu \in ca(\Sigma ,X)$, then C(K) is isomorphic to $c_0$.

Proof

Such C(K) space is WCD, because every subspace of a weakly compactly generated Banach space (like $L^1(\nu )$) is WCD. Proposition 3.5 applies to deduce that C(K) is separable, i.e. K is metrizable. On the other hand, since every subspace of an order continuous Banach lattice (like $L^1(\nu )$) has the so-called Pełczyński’s property (u) (see e.g. [2, Theorems 4.54 and 4.56]), so does C(K). It follows that C(K) is isomorphic to $c_0$ (see e.g. [1, Theorem 4.5.2]). $\square$