Let $M([0,1])$ be the set of finite signed measures on $[0,1]$
(with the topology generated by the sets $\left\{ \mu \in M([0,1]) : \left| \int f(x) \mu(dx)- a\right| \leq \delta\right\}$ for all $\delta>0$, $a \in R$ and $f \in C\_b([0,1])$ (continuous and bounded). (hence weak-\*-topology)


[Here](https://math.stackexchange.com/q/700545/246867) it is answered that $M([0,1])$ is not first countable.


But is it also not sequential; that is, does sequential continuity of a function $F : M([0,1]) \rightarrow \mathbb R$ imply continuity of $F$ in general or is that not the case?



Is the space of signed finite measures on a compact set $M([0,1])$ a sequential space?
It is not sequential. Here is an explicit counterexample.


Notation: $\|\mu\|$ is the total variation norm of $\mu$, and $\bar{B}\_r \subset M([0,1])$ is the closed norm ball of radius $r$ centered at $0$. $\delta\_x$ is the Dirac measure which places a unit point mass at $x \in [0,1]$.


For each positive integer $n > 0$, let $P\_n \subset [0,1]$ be any finite set of size at least $(2n)^{2n}$, and let
$$E\_n = \{ n(\delta\_{x} - \delta\_y) : x,y \in P\_n, \, x \ne y\} \subset M([0,1]).$$


Notice that $E\_n$ is finite, and that $\|\mu\| = 2n$ for every $\mu \in E\_n$.


Set $E = \bigcup\_{n=1}^{\infty} E\_n$. I claim that 0 is a weak-\* limit point of $E$. Let $f\_1, \dots, f\_m \in C([0,1])$ and let $\epsilon > 0$. I will produce $\mu \in E$ with $\left|\int f\_k\,d\mu\right| < \epsilon$ for every $k$. Without loss of generality, assume $\|f\_k\|\_\infty \le 1$ for all $k$ (otherwise divide $\epsilon$ by $\max\_k \|f\_k\|\_\infty$). Define $F : [0,1] \to [-1,1]^m$ by $F = (f\_1, \dots, f\_m)$. Choose $n > \max(m, 1/\epsilon)$. 


We can cover $[-1,1]^m$ with $(2n^2)^{m}$ cubes of side length $1/n^2$. Since $$|P\_n| \ge (2n)^{2n} > (2n)^{2m} = (4n^2)^m > (2n^2)^m$$ by the pigeonhole principle there must exist two distinct $x,y \in P\_n$ with $F(x), F(y)$ in the same cube. This means that $|f\_k(x) - f\_k(y)| \le 1/n^2$ for every $k$. So if we take $\mu = n(\delta\_x - \delta\_y) \in E\_n$, we have $\left|\int f\_k\,d\mu\right| \le 1/n < \epsilon$ for every $k$. This proves the claim that 0 is a weak-\* limit point of $E$.


Now for any $n$, we know that $\bar{B}\_n$ is weak-\* closed and disjoint from the finite set $E\_n$. [The weak-\* topology is completely regular](https://math.stackexchange.com/questions/352661/is-the-weak-star-topology-on-the-dual-of-a-banach-space-completely-regular), so there exists a weak-\* continuous function $G\_n : M([0,1]) \to [0,1]$ with $G\_n = 0$ on $\bar{B}\_n$ and $G\_n = 1$ on $E\_n$. Set $G = \sum\_{n=1}^\infty G\_n$. Note that on any ball $\bar{B}\_n$, only finitely many terms of the sum are nonzero, so the sum makes sense, and the restriction of $G$ to any ball is weak-\* continuous. If $\mu\_k$ is a weak-\* convergent sequence with limit $\mu$, then by the uniform boundedness principle all the $\mu\_k$ and $\mu$ lie in some ball $B$. The restriction of $G$ to $B$ is weak-\* continuous, so $G(\mu\_k) \to G(\mu)$. Thus we have shown $G$ is weak-\* sequentially continuous.


On the other hand, $G$ is not weak-\* continuous, since $G(0) = 0$ but $G \ge 1$ on $E$.


(The same uniform boundedness argument shows that $E$ is weak-\* sequentially closed, but not weak-\* closed since it does not contain 0.)


More generally, we can replace $M([0,1])$ by the dual of any separable Banach space. See the paper:



> 
> Humphrey, A. James and Simpson, Stephen G. Separable Banach space theory needs strong set existence axioms. *Trans. Amer. Math. Soc.* **348** (10), 4231-4255, 1996. [Open access full text](http://www.ams.org/journals/tran/1996-348-10/S0002-9947-96-01725-4/S0002-9947-96-01725-4.pdf)
> 
> 
> 


Theorem 2.5 of that paper shows that for any infinite-dimensional separable Banach space $X$, there is a countable subset $Z \subset X^\*$ which is weak-\* sequentially closed but weak-\* dense. In particular, $Z$ is not weak-\* closed, so the weak-\* topology on $X^\*$ is not sequential in the usual sense of the word. 


To show it is not sequential in your sense: As an intermediate step of their construction, they get a sequence which has $0$ as a weak-\* limit point but intersects every ball in only finitely many points; we could proceed as we did above to construct a function $G : X^\* \to \mathbb{R}$ which is weak-\* sequentially continuous but not weak-\* continuous. (Their set $Z$ is formed by translating such a sequence by a countable weak-\* dense set.)


