Let $X$ be a topological space and $Y\subseteq X$, the sequential closure of $Y$ is the set of elements in $X$ that are limit of **sequences** belonging to $Y$.


Let $\mathcal M\_{\text{fin}}(\mathbb Z)$ be the set of finitely supported probability measures on $\mathbb Z$. The general question is: who is the weak\* sequential closure of $\mathcal M\_{\text{fin}}(\mathbb Z)$? This set should be different from the set of all finitely additive probability measures on $\mathbb Z$, since $L^\infty(\mathbb Z)$ is not separable. 


I'm actually interested in a more specific question: does this closure contain at least one (all??) translation invariant finitely additive probability measure?


Thanks in advance,


Valerio



Who is the weak* sequential closure of the set of finitely supported measures on the integers?
The finitely support probability measures on $\mathbb Z$ are all members of $\ell^1(\mathbb Z)$. So we could ask a slightly more general question:



> 
> What is the sequential closure of (the probability measures in) $\ell^1(\mathbb Z)$ in $\ell^\infty(\mathbb Z)^\*$?
> 
> 
> 


If $(a\_n)$ is a sequence in $\ell^1(\mathbb Z)$ which converges weak$^\*$ in $\ell^\infty(\mathbb Z)^\*$, then $f(a\_n)$ is a Cauchy sequence for all $f\in \ell^\infty(\mathbb Z)$. Now, $\ell^1(\mathbb Z)$ is weakly sequentially complete (\*) and so there is $a\in\ell^1(\mathbb Z)$ with $f(a\_n) \rightarrow f(a)$ for each $f\in\ell^\infty(\mathbb Z)$. In other words, the weak$^\*$ limit of $(a\_n)$ actually lives in $\ell^1(\mathbb Z)$. So the answer is "no" in a strong sense.


(\*) All preduals of a von Neumann algebra are weakly sequentially complete (see Takesaki I, Chapter III, Corollary 5.2). There is presumably an easy proof for $\ell^1$, but I don't know a reference off the top of my head.


**Edit:** The following is prompted by Bill Johnson's comment (and reveals nothing so much as my own ignorance!) A Banach space $E$ is said to have the Schur Property if weakly convergent sequences are norm convergent-- Schur originally showed that $\ell^1$ has the Schur Property. We say that $E$ is weakly sequentially complete (WSC) if whenever $(x\_n)$ is weakly Cauchy (meaning that for each $f\in E^\*$, the scalar sequence $(f(x\_n))$ is Cauchy) then $(x\_n)$ is weakly convergent. The Schur property implies WSC-- if $(x\_n)$ is weakly Cauchy, then for any increasing sequences $(n\_k)$ and $(m\_k)$, the sequence $(x\_{n\_k} - x\_{m\_k})\_k$ is weakly-null, hence norm-null, and hence $(x\_n)$ is norm Cauchy. You can find all this in Section 2.3 of Albiac and Kalton's book. For a general measure $\mu$, the space $L^1(\mu)$ is WSC, but doesn't necessarily have the Schur property-- see Section 5.2 of this book.


