A question on the dual of $l_\infty(X)$ Let $X$ be a compact Hausdorff space. Let $C(X)$ be the continuous functions on $X$ with sup norm and $l_\infty(X)$ be the bounded functions on $X$, also with the sup norm.
Suppose $\{g_n\}_{n\in\mathbb{N}}\subset C(X)$ is such that $g_n\to g$ pointwise, whence $g\in l_\infty(X)$.
Edit: I am assuming $\{g_n\}$ is uniformly bounded here.
Does $g_n \to g$ weakly in $l_\infty(X)$?
The motivation for this question is as follows:
For any functional $T\in l_\infty(X)^*$, $T\mid_{C(X)}$ corresponds to a Radon measure $\mu$ by the Riesz representation theorem. Thus, by DCT,
$$T(g_n) = \int g_n(x) d\mu(x) \to \int g(x) d\mu(x)$$
However, can we conclude that $\int g(x) d\mu(x) = T(g)$? 
(What if we do this on Bounded Borel functions instead of $l_\infty(X)$?)
Thanks!
 A: First off, you cannot deduce that
$$(*)\qquad T(g_n)=\int_X g_nd\mu\to \int_X g d\mu $$
by the DCT under the sole assumption of $g_n\to g$ pointwise. Take for instance $X=[0,1]$ and $\mu$ the Lebesgue measure - you can easily construct a sequence of continuous functions which converges pointwise to $g=0$, but whose integrals do not converge to $0$.
You also need 
$$\exists f\in L^1(X):|g_n(x)|\leq |f(x)|\quad\text{for a.e.}\,x\in X,\;\forall n\in \mathbb{N} $$
This is for instance guaranteed if $\left\{g_n\right\}$ is uniformly bounded, i.e.  there is some constant $C>0$ such that $\|g_n\|_{\infty}\leq C$ for all $n\in \mathbb{N}$.
Secondly, the answer to your first question is no.
Indeed, choose a sequence of distinct points $\left\{x_n\right\}_{n\in \mathbb{N}}\subset X$ and take 
\begin{align*}F(g):=\sum_{n=1}^{\infty}g(x_n)2^{-n}
\end{align*}
The series converges for all $g$ bounded, and $F$ is linear and continuous on $\ell_{\infty}$, because if $\|g_k\|_{\infty}\to 0$ then
$$|F(g_k)|\leq \|g_k\|_{\infty}\sum_{n=1}^{\infty}2^{-n}=\|g_k\|_{\infty}\to 0 $$ Now since $X$ is Hausdorff, let $\left\{U_n\right\}$ be a sequence of disjoint neighborhoods of $\left\{x_n\right\}$ and take a sequence of continuous functions $\left\{g_n\right\}$, satisfying the following properties: 
\begin{align*}g_n(x)&=0 \qquad \forall x\notin U_n \\
g_n(x_n)&=2^n\qquad \forall n\in \mathbb{N} \end{align*}
(such a sequence exists by the Urysohn lemma since all compact Hausdorff spaces are normal).
Then for each $x\in X$, we have two options. Either $x\notin \bigcup_{n\in\mathbb{N}}U_n$, and then $g_n(x)=0$ for all $n$, or $x\in U_m$ for a unique $m\in \mathbb{N}$, and then $g_n(x)=0$ for all $n\neq m$. In both cases, $g_n(x)\to 0$, so $g_n\to 0$ pointwise. However,
$$ F(g_n)=2^n\cdot2^{-n}=1\not \to 0$$
and hence $g_n\not \to 0$ weakly.
EDIT: However, if you additionally assume that $\left\{g_n\right\}$ is uniformly bounded in $C(X)$, then weak convergence in $\ell_{\infty}(X)$ does hold.
Indeed, under this assumption, by the dominated convergence theorem we have 
$$\int_X g_n\,d\mu \to \int_X g\,d\mu $$
for all Radon measures $\mu$. By the Riesz Representation theorem, this means that for each $T\in l_{\infty}(X)^*$, the sequence $\left\{T(g_n)\right\}$ is convergent, and hence $\left\{g_n\right\}$ is weakly convergent in $l_{\infty}(X)$. It remains to prove that the weak limit is $g$. Suppose that $g_n\rightharpoonup h$, with $h\in l_{\infty}(X)$. Then for each $x\in X$, if $T_x\in l_{\infty}(X)$ is the evaluation functional
\begin{align*}T_x:l_{\infty}(X)&\to \mathbb{R}\\ 
f&\mapsto f(x)
\end{align*}
since $g_n\rightharpoonup h$ we have $T_x(g_n)\to T_x(h)$ for all $x\in X$, i.e. $g_n(x)\to h(x)$ for all $x\in X$. On the other hand since $g_n\to g$ pointwise we have $g_n(x)\to g(x)$ for all $x\in X$. Thus $g=h$, and $g_n\rightharpoonup g$ weakly. In particular, $T(g_n)\to T(g)$ for all $T\in l_{\infty}(X)^*$ and
$$T(g)=\int_X g\,d\mu $$
