Questions on topologies on space of Radon measures

Question

Consider the space $C_c(\mathbb{R})$ of continuous real-valued functions on $\mathbb{R}$ equipped with the inductive limit topology by $C_c(\mathbb{R}) = \bigcup_{n \in \mathbb{N}} C_c(\mathbb{R}, K_n)$ where $K_n$ is some compact exhaustion of $\mathbb{R}$ by compact sets and $C_c(\mathbb{R}, K_n)$ is the Banach space with uniform norm. In particular, $C_c(\mathbb{R})$ is an LB-space. Let $M := M(\mathbb{R})$ denote the topological dual $C_c'(\mathbb{R})$. Then $M$ can be identified with the space of all Radon measures (bounded and unbounded, e.g. the Lebesgue measure belongs to $M$) - for details, see Bourbaki: Integration, Chapter III. Every measure $\mu \in M$ has the Jordan decomposition $\mu = \mu^+ - \mu^-$ into two positive measures $\mu^+$ and $\mu^-$. Let $M^+ \subseteq M$ denote the subset of all positive measures. Then we have a surjective map $s : M^+ \times M^+ \to M$, $(\mu, \nu) \mapsto \mu - \nu$.

One can equip $M$ with the weak-* topology $\tau_v$ which Bourbaki calls the vague topology.

Some properties of $(M, \tau_v)$:

• $M$ is by duality the projective limit of the weak-* duals $C_c'(\mathbb{R}, K_n)$.
• $M$ is Hausdorff locally convex and in particular, addition and scalar multiplication are continuous
• $M$ is not first-countable and thus not metrizable
• $M$ is quasi-complete but not complete
• The map $s$ is continuous ($M^+$ carries the subspace topology (which is Polish))
• $M$ is Souslin (as the image of the Polish space $M^+ \times M^+$ under the continuous map $s$)
• $M$ is separable (since Souslin)

I don't know of the following:

• Is $M$ sequential or even Fréchet-Urysohn? (This is a specialization of the question here.)
• Is $M$ Lusin? (For an uncertain proof idea see below.)

Bourbaki studies also other topologies like compact convergence or strictly compact convergence. But one has also the map $s$ from above and can equip $M$ with the final topology $\tau_f$ induced by $s$ which is then finer than $\tau_v$ and turns $M$ into a quotient space. In particular, $(M, \tau_f)$ is sequential. Does anyone know whether this topology on $M$ was already studied before and has some nice properties (or missing some good properties)? In particular, is $\tau_f$ a vector space topology (addition and scalar multiplication continuous)?

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Proof for $(M, \tau_v)$ is Lusin: I'm trying to transfer the ideas of Trèves, "Topological Vector Spaces", p. 556 (in particular Proposition A.9) which is based on the Borel Graph Theorem. $(M, \tau_v)$ is the projective limit of the countable family $C_c'(\mathbb{R}, K_n)$ equipped with the weak-* topology $\tau_{w*}$. The space $C_c'(\mathbb{R}, K_n)$ equipped with strong norm topology is Banach but it is not separable, thus not Polish and thus we can't directly say whether $(C_c'(\mathbb{R}, K_n), \tau_{w*})$ is Lusin. Now fix $n$ and set $E := C_c(\mathbb{R}, K_n)$. Since $E$ is separable Banach we have a countable basis $U_k$ of open nbhds. of $0$ and the dual $E'$ is the union of all the polars $U_k^0$. Each $U_k^0$ is weakly compact (Banach-Alaoglu) and thus weakly closed equicontinuous. By Trèves Exc. 32.9 it follows that $U_k^0$ equipped with the weak-* topology is metrizable compact, thus Polish and therefore Lusin. A countable union of Lusin subsets of a Hausdorff space is Lusin and thus $E'$ with the weak-* topology is Lusin. In other words, $(C_c'(\mathbb{R}, K_n), \tau_{w*})$ is Lusin. Moreover, the product of the countable family $(C_c'(\mathbb{R}, K_n), \tau_{w*})$ of Lusin spaces is Lusin and the projective limit $M$ of all the $(C_c'(\mathbb{R}, K_n), \tau_{w*})$ is a closed subset of the product thus a Borel subset of the product and thus Lusin.

The only transfer of the proof is that I replaced "Souslin space" by "Lusin space". So is there some major gap which I did oversee?

Answer 1

It's not sequential because its closed subspace $M[0,1] = (C[0,1])^\ast$ is not sequential.

Here is an example of a set $A \subset M[0,1]$ that is sequentially $\tau_v$-closed but not $\tau_v$-closed:

Consider sequence of functions $f_n \in C[0,1]$, $\Vert f \Vert_{C[0,1]} = 1$ and $\operatorname{span} \{f_n\}$ is norm-dense in $C[0,1]$, and take $$A := \bigcup_{n \ge 1} \left\{ \mu : \intop f_n d \mu = n \right\},$$

On the one hand, it's sequentially $\tau_v$-closed. Indeed, any $\tau_v$-convergent sequence is bounded in the total variation norm by the uniform boundedness principle, and $A$ intersects any ball at only finitely many closed hyperplanes. On the other hand, $0 \notin A$ but $0$ is contained in the $\tau_v$-closure of $A$ because any basic $\tau_v$-neighborhood of $0$ of the form $\left\{ \left| \intop g_i d \mu \right| < \varepsilon, i = 1, \dots, n \right\}$ intersects $\left\{ \intop f_m d \mu = m \right\}$ for any $f_m \notin \operatorname{span} \{ g_i\}$.

It's Lusin.

One can "encode" a measure on $\mathbb{R}$ by a sequence of compatible finite measures on $[-n, n]$. Now, on finite measures on $[-n, n]$ the weak topology can be strenghthened to a Polish space topology by adding total variation norm balls around $0$ as additional open sets. Here we are using the fact that on balls the weak topology becomes metrizable. For example, one can use the following metric: $$\rho(\mu, \nu) := \sum_m \frac{1}{2^m} \arctan \left| \intop f_m d (\mu - \nu) \right| + \left| \Vert \mu \Vert - \Vert \nu \Vert \right|$$ where $\Vert \cdot \Vert$ is the total variation norm and $f_m$ is a dense sequence of functions of norm $1$.

The condition that the measures on $[-n, n]$ are compatible introduces a closed subset in the countable product of spaces of measures on $[-n, n]$. So this way we can cook up a Polish space topology on $M(\mathbb{R})$.

Answer 2

This is a comment rather than an answer but it will be too long. It mentions some tools that might be of use. To begin with, there are advantages in using the context of a general locally compact space $X$ rather than the special case of the real line, at least in the beginning.

1. In the case of a compact space $K$, there are three natural topologies on $M(K)$, the weak topology induced by $C(K)$, the norm topology and a third less familiar one, the bounded weak star topology which is the finest locally convex topology which agres with the weak topology on the unit ball. This has several advantages over the other two. In particular, it is complete as the weak topology is not. However, it has the same convergent sequences and coincides with it on bounded sets, importantly on the set of probability measures.

2. As you point out, the dual space in the general case is a projective limit of the duals of analogues of those above, namely of Banach spaces of those functions with supports in a fixed compact subset. It is natural to consider it as a (complete) locally convex space with the corresponding limit structure. This will not be identical to the weak topology you use but will be closely related.

3. The space of measure will then be naturally isomorphic to a closed subspace of a product of spaces of measures with compact suport, as mentioned above. In the case of a $\sigma$-compact space this will be a countable product.

4. Hence it will inherit topological properties of the space of measures on compacta (in your case, even compact metric spaces) which are stable under (countable) products and closed subspaces.

5. How ever, in the case where $X$ is paracompct (and so, in particular, $\sigma$-compact), it is even a complemented subspace and thus a continuous image of the product. This follows from results of de Wilde and Keim on projective limits with partitions of unity which can be found in Manuscripta Mathematica, vols. 5 and 10.

I have no expertise in descriptive topology and can't say whether this helps in your problem but it might be worth your while to have a look.