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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.

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    $\begingroup$ First prove this for hypercubes $\Omega = \prod_{i=1}^d (a_i, b_i)$, then try to generalise using that every bounded open set in $\mathbb R^d$ can be exhausted by open hypercubes. $\endgroup$
    – AlexR
    Dec 18 '14 at 21:18
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$\Rightarrow$: 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.

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    $\begingroup$ Worth mentioning in $\impliedby$ that boundedness is used to carry through the approximation. $\endgroup$
    – fourierwho
    Jul 22 '17 at 0:07

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