According to the book Dunford-Schwartz, Linear Operators I, Theorem IV.8.16 the space $ (L^{\infty} (S, \Sigma, \nu))^*$ is isomorphic the space $ba (S, \Sigma_1, \| . \|)$. Unfortunately, I ran into the problem that I don't understand how the isomorphic mapping works. In Dunford-Schwartz' book the isomorphism is shown as follows: $$ F (l) = \int_{S} l(s) \lambda (ds), $$ where $F \in L^{\infty} (S, \Sigma, \nu)^*, l \in L^{\infty} (S, \Sigma, \nu), \lambda \in ba $. Moreover we know the relationship between $\lambda$ and $F$: $$ \lambda (E) = F ( I_E ), ~ \text{for any} E \in \Sigma $$ where $I_E$ characteristic function of the set: $$ I_E ( a ) = \left\{ \begin{aligned} &1, ~ \text{if } a \in E \\ &0, ~ \text{if } a \notin E \end{aligned} \right. $$ Let's look at the set $l^{\infty} = L^{\infty} ( \mathbb{N}, 2^{\mathbb{N}} )$ with standart counting metric. Now I can tell you what my question is. It is known $l^{\infty}$ is not reflexive in particular , there is $ \phi \in (l^{\infty})^*$ such that $ \phi (c) = 0 $, where $c$ is the space of sequences having a finite limit and $ \| \phi \| = 1 $. But for any $E \in 2^{\mathbb{N} }$ $ I_E $ is sequences having a finite limit it means that $\lambda(E) = 0$, but $\phi$ is not identically a null function.
I will be happy to see an explanation of how the said isomorphism works.
Edit: I made a mistake when describing the question. But unfortunately I haven't fully understood out this problem yet.
Metric in space $ba$ is total variation.It's easy to make sure $l_1 \subset ba$. Do I understand correctly that $$ F(l) = \sum_{n=1}^{\infty} l(n) \lambda( \{ n \} ) $$ where $F \in (l^{\infty})^*, \lambda \in ba$ ?
In this case, we will choose $ l(n) = sign ( \lambda \{n\} ), \| l \| = 1 $ and $$ F(l) = \sum_{n=1}^{\infty} | \lambda( \{ n \} ) | \leq \| F \| $$ it means $ \{ \lambda(\{ n \} ) \}_{n=1}^{\infty} \in l^1 $ but $l^{1} $ is not reflexive.