Prove that $f \in \overline{ \{e_n \} }^{\omega^*} \subset l_\infty^* \iff f(1_A)=1 \ or \ 0 $ I've been trying to prove this claim for a long time ago:

$f \in l_\infty ^*$ assigns 0 or 1 over the characteristic functions
  (understood as sequences of 0's and 1's) if and only if $f \in \overline{ \{e_n \} }^{\omega^*}$, where $\omega^*$ is the weak* topology over $l_\infty^*$ and $\{e_n \}$ is the basis of $l_1$ (seen
  in $l_1^{**}=l_\infty ^*$).

I have been able to prove the converse implication ($\Leftarrow$), but I can't do the direct implication.
 A: One of the implications can be proved, for example, using nets.
If $f$ belongs to the closure of $\{e_n\}$, then there exists a net $(e_{n_d})_{d\in D}$ which converges to $f$ (in weak${}^*$-topology).
The convergence of a net in the weak${}^*$ topology means that 
$$(\forall x\in\ell_\infty) \lim_{d\in D} e_{n_d}(x)=f(x).$$ 
Specifically we get that 
$$f(\chi_A)=\lim_{d\in D} e_{n_d}(\chi_A).$$
But on the RHS we have a net of real numbers from the set $\{0,1\}$. Such net can only convergent to an element of $\overline{\{0,1\}}=\{0,1\}$. So we see that $f(\chi_A)$ is either $0$ or $1$.
We may also notice that, $f(\chi_{\mathbb N})=1$, since $e_n(\chi_{\mathbb N})=1$ for each $n$. Thus $f\ne 0$.
Alternatively we could use ultralimits instead of nets. (Which would probably made the whole answer look more unified, since ultralimits are used in the second, more difficult, part. The answer is written in this way because I originally misunderstood which implication is the OP asking for. But perhaps this also has advantages - I guess that nets are known to more people than ultrafilter limits, which makes at least the easier part more accessible.)

Now let us try to prove the other implication.
Let $f\in\ell^*_\infty$ be a non-zero function such that the values $f(\chi_A)$ are zeroes and ones. We want to show that $f$ belongs to closure of $\{e_n\}$ in the weak${}^*$-topology.
To prove this, we will prove following things.


*

*Fact 1. $f$ is a positive functional, i.e., for $x\in\ell_\infty$ we have
$$x\ge 0 \implies f(x)\ge 0.$$
Notice that this also gives that $f(x)\le f(y)$ whenever $x\le y$.

*Fact 2. $f(\chi_{\mathbb N})=1$ and $\|f\|=1$.

*Fact 3. The set 
$$\mathfrak p=\{A\subseteq\mathbb N; f(A)=1\}$$ is an ultrafilter.

*Fact 4. In the weak${}^*$-topology we have
$$\newcommand{\plim}{\operatorname{{\mathfrak p}-lim}} \plim e_n = f.$$


Fact 4 implies that $f$ belongs to the closure of $\{e_n; n\in\mathbb N\}$.

Proof of Fact 1. Notice that if $x\ge 0$ then $x$ can be approximated (in $\ell_\infty$ norm) by a sequence of step sequences1
$$s^{(n)}=\sum_{i\in F^{(n)}} c_i \chi_{A_i}$$
where $F^{(n)}$ is a finite index set, $\{A_i; i\in F^{(n)}\}$ is a system of disjoint subsets of $\mathbb N$ and $c_i\ge 0$. (And we have similar approximation for arbitrary $x\in\ell_\infty$, if we omit the condition $c_i\ge 0$. This is related to another your question.) 
So now we have $$f(s^{(n)}) = \sum c_i f(\chi_{A_i}) \ge 0.$$
Since $s^{(n)}\to x$ and $f$ is continuous, we also have $f(x)\ge0.$

Proof of Fact 2. If $f(\chi_A)=0$ for each $A$, then $f(s)=0$ for each step sequence of the form $s=\sum\limits_{i\in F} c_i \chi_{A_i}$. Since such sequences are dense in $\ell_\infty$, we would get $f=0$, a contradiction.
So $f(\chi_A)=1$ at least for some $A\subseteq\mathbb N$. From the monotonicity we get also $f(\chi_{\mathbb N})=1$.
Now if $\|x\|\le 1$, this means that $-\chi_{\mathbb N} \le x \le \chi_{\mathbb N}$ and, using monotonicity again,
$$-1 = f(-\chi_{\mathbb N}) \le f(x) \le f(-\chi_{\mathbb N}) =1.$$
This implies that $\|f\|=1$.

Prof of Fact 3. We already have $\mathbb N\in\mathfrak p$ and $\emptyset\notin\mathfrak p$.
Now if $A\in\mathfrak p$ and $A\subseteq B$, then we have $\chi_A\le \chi_B$. So monotonicity implies that
$$1=f(\chi_A)\le f(\chi_B)$$
and since $f(\chi_B)$ can only be zero or one, we get that $f(\chi_B)=1$ and $B\in\mathfrak p$.
Let $A,B\in\mathfrak p$. Then we have
\begin{align*}
1=\chi_A&= \chi_{A\cap B}+\chi_{A\setminus B}\\
1=\chi_B&= \chi_{A\cap B}+\chi_{B\setminus A}
\end{align*}
Since the values on the RHS are only zeroes and ones, we have only two possibilities: Either $\chi_{A\cap B}=1$ and $\chi_{A\setminus B}=\chi_{B\setminus A}=0$ or $\chi_{A\cap B}=0$ and $\chi_{A\setminus B}=\chi_{B\setminus A}=1$. But the latter is impossible, since it would imply that $2=\chi_{A\setminus B}+\chi_{B\setminus A}\le \chi_{\mathbb N}=1$. So we get $\chi_{A\cap B}=1$ and $A\cap B=\mathfrak p$.

Prof of Fact 4. Since unit ball $B$ of $\ell_\infty^*$ is compact in weak${}^*$-topology by Banach-Alaoglu theorem, the ultafilter limit $\plim e_n$ exists and belongs to $B$. (We are using this basic fact about ultralimits of sequences in compact spaces.)  Let us denote $p=\plim e_n$.
So we have that both $p$ and $f$ are bounded linear functional belonging to $B$. We want to show that $\plim e_n(x)=f(x)$ for each $x\in\ell_\infty$.
We first show that these two functionals have the same values for $x=\chi_A$. Indeed, we have
$$p(\chi_A)=\plim e_n(\chi_A) = \plim \chi_A(n) =
  \begin{cases}
    1 & A\in\mathfrak p, \\
    0 & A\notin\mathfrak p,
  \end{cases}
$$
which means $\plim e_n(\chi_A)=f(\chi_A)$ for each $A\subseteq\mathbb N$. (Notice that we have used the fact that $\mathfrak p$-convergence in weak${}^*$-topology is equivalent to pointwise $\mathfrak p$-convergence.)
Now this implies also $p(\sum c_i\chi_{A_i})=f(\sum c_i\chi_{A_i})$ for any finite linear combination. So we have that $p$ and $f$ coincide on a dense subset of $\ell_\infty$ and thus $p=f$.

1 Step sequence is the name I use for finite linear combinations of characteristic sequences. I do not know if there is a standard terminology of this - probably I could use the names step function and characteristic function. After all, sequences are simply functions defined on $\mathbb N$. We can also see that step sequences are precisely those elements from $\ell_\infty$ which have only finitely many values.
