If $(\Omega, \Sigma, \mu)$ is a (complete) $\sigma$-finite measure space then $(L^{\infty}(\Omega,\Sigma,\mu))^{\ast}$ is the space $\operatorname{ba}(\Omega, \Sigma,\mu)$ of all finitely additive finite signed measures defined on $\Sigma$, which are absolutely continuous with respect to $\mu$, equipped with the total variation norm.
In particular the dual of $\ell^{\infty}=\ell^{\infty}(\Bbb N)$ equals to $\operatorname{ba}(\Bbb N,p(\Bbb N),\sharp),$ where $\sharp$ denots the counting measure. Now, is the relationship $\big(\ell^{\infty}(X)\big)^*= \operatorname{ba}(X,p(X),\sharp)$ holds? (Although, $(X,p(X),\sharp)$ is not a $\sigma$-finite measure space when $X$ is assumed to be an uncountable set)
If $F \in (l^\infty(X))^*$, define $\mu \in \operatorname{ba}(X,p(X),\sharp)$ by $$ \mu(A) = F(\mathbf1_A),\quad A \subseteq X . $$ Three things:
(a) Is $\mu$ really in $\operatorname{ba}(X,p(X),\sharp)$? Seems easy.
(b) Is $F \mapsto \mu$ injective? That is: if $F_1(\mathbf1_A)=0$ for all $A \subseteq X$, then $F(f) = 0$ for all $f \in l^\infty(X)$. Seems clear: any $f \in l^\infty(X)$ is the uniform limit of simple functions.
(c) Is $F \mapsto \mu$ surjective? Given $\mu \in \operatorname{ba}(X,p(X),\sharp)$, construct the corresponding $F \in l^\infty(X)$. There is more work here, but shouldn't this be the same as the proof for $X = \mathbb N$? First define $F$ on simple functions. If simple functions $s_n$ converge uniformly to $0$, then $\lim_n F(s_n) = 0$. Given $f \in l^\infty(X)$, there exist simple functions $s_n$ converging uniformly to $f$; show $(F(s_n))_n$ is Cauchy, then define $F(f) = \lim_n F(s_n)$.