# Radon-Nikodym property for space of signed measures

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.

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.

• 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? Apr 24, 2021 at 20:07
• @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. 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.