TITLE: weak* convergence for sequence in $ L^\infty$
QUESTION [5 upvotes]: Let $ \Omega \subset \mathbb{R}^d $ be a bounded and open set. Suppose $ \{f_n\} \subset L^{\infty} (\Omega) , f \in L^{\infty} (\Omega) $. 
Prove that
$ f_n \rightharpoonup^* f \ \ \text{in} \ \  L_{\infty} (\Omega)  \Longleftrightarrow \left(   \sup\limits_{n \in \mathbb{N} } \| f_n \|_{L^{\infty}} < + \infty \ \ \text{and} \ \ { \displaystyle \forall_{ V \subset \Omega} \ \lim\limits_{n \to \infty} \ \int\limits_V \ (f_n(x) - f(x)) =0} \right)   \\ \text{where } V= \Pi^d_{i=1}(a_i,b_i). $
Can anyone give me any clue? I would appreciate any help.

REPLY [2 votes]: $\Rightarrow$: a weak* convergent sequence is bounded. For the remaining part, use the definition of weak* convergence with the characteristic function of $V$. 
$\Leftarrow$: if $g$ is integrable on $\Omega$, we can approximate in $\mathbb L^1(\Omega)$ this function by linear combinations of sums of charactistic functions of cubes.