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](https://math.stackexchange.com/questions/1465574/are-lusin-and-souslin-spaces-sequential-or-even-fr%C3%A9chet-urysohn).)
* 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?



Questions on topologies on space of Radon measures

> 
> 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 $M(\mathbb{R})$.


