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 $M(\mathbb{R})$.

The quotient topology for the map $M^+ \times M^+ \to M$ is not a vector space topology.

The idea is that the neighnorhoods of $0$ are those of the local total variation topology, but overall the topology is weaker than total variation, so the space is not even homogeneous w.r.t. shifts.

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 $M(\mathbb{R})$.

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 $M(\mathbb{R})$.

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 $M(\mathbb{R})$.

The quotient topology for the map $M^+ \times M^+ \to M$ is not a vector space topology.

The idea is that the neighnorhoods of $0$ are those of the local total variation topology, but overall the topology is weaker than total variation, so the space is not even homogeneous w.r.t. shifts.