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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 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?

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 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?

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 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?

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 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?