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)




This implies that for each $f \in C_b([0,1]\times [0,1],\mathbb R)$ (continuous and bounded) the set
$$
  A=\left\{ \mu \in M([0,1]) : \int_{[0,1]} \int_{[0,1]} f(x,y) \mu(dx) \mu(dy)
   \leq
    1
  \right\} 
$$
is sequentially closed.


[Here](http://mathoverflow.net/posts/208983/edit) it is answered that $M([0,1])$ is not sequential.
Therefore we can not infer from the sequentially closedness of $A$ that it is also closed.



Question: Is $A$ nevertheless closed?