Given a measurable space, the vector space of signed measures is a Banach space. Does it have the Radon-Nikodym property? What if the space is of a special type, such as a nice topological space with the Borel $\sigma$-algebra?

If there is any related information, such as the Radon-Nikodym property relative to some special measure spaces, I would also be interested.

I assume the answer is well known, but I have not been able to find information in standard sources such as Diestel and Uhl's book.


3 Answers 3


The spaces you are interested in are abstractly AL-spaces and by Kakutani's representation theorem, they can be represented as $L_1(\mu)$ for some measure. In particular, they have the RNP if and only if the measure $\mu$ is purely discrete, in which case $L_1(\mu)$ is isometric to $\ell_1({\rm supp}\, \mu)$.

Thus, for compact spaces $K$, the space of measures $M(K)$ has RNP if and only if $K$ is scattered.


This is more or less the classical example of a space without RNP. For example, in the case of the unit interval with the Borel algebra, it is a non separable dual of a separable Banach space and so fails RNP. More general situations can be dealt with using Stegall’s theorem that the dual of a Banach space $E$ has RNP if and only if every separable subspace thereof has a separable dual, a result which is certainly in Diestel and Uhl.

  • $\begingroup$ I suppose I'm missing some elementary facts. I know that the space of finitely additive signed measures is a dual space. But what is the predual of the space of countably additive signed measures? $\endgroup$ Commented Apr 24, 2021 at 20:07
  • 1
    $\begingroup$ @QuartoBendir: The signed Borel measures on $[0,1]$ are the dual space of $C([0,1])$ (the space of continuous real-valued functions). That's a classical representation theorem. $\endgroup$ Commented Apr 24, 2021 at 22:04

The problem is already answered above. Perhaps it is worth to note the natural "next steps" one would take after this.

If $K$ is a compact topological space and $M(K) = C(K)^{\ast}$, then the following are equivalent.

  1. $M(K)$ has RNP.
  2. $C(K)$ does not contain a copy of $\ell^1$.
  3. $M(K)$ has the Schur property. (weakly convergent sequences are norm convergent)

$(1\Leftrightarrow 2)$ is already answered above. $(2\Leftrightarrow 3)$ for any Banach space $X$, $X^{\ast}$ has Schur property iff $X$ has Dunford-Pettis property (DPP) and contains no copy of $\ell^1$. $C(K)$ has DPP.

Next step is to generalize this from $C(K)$ to a general $C^{\ast}$-algebra $A$.

(a) the following are equivalent for $A$: (see Chu, Huruya, and Jensen for the implication $(4\Rightarrow 1)$)

  1. $A^{\ast}$ has RNP.
  2. $A$ does not contain a copy of $\ell^1$.
  3. $A$ does not contain a copy of $C([0,1])$.
  4. If $x\in A$ is self-adjoint, then its spectrum $\sigma(x)$ is countable. (for otherwise $C(\sigma(x))$ and $C([0,1])$ are isomorphic.)

(b) the following are equivalent for $A$ by Hamana:

  1. $A^{\ast}$ has DPP.
  2. $A$ has DPP
  3. Every irreducible representation of $A$ is finite dimensional.

(c) Thus by (a)&(b), the following are equivalent for a $C^{\ast}$-algebra $A$.

  1. $A^{\ast}$ has the Schur property.
  2. $A^{\ast}$ has DPP and RNP.

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