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Let $\mathcal{F}$ be a bounded set in $L^{1}(\Omega)$. Then $\mathcal{F}$ has compact closure in the weak topology $\sigma(L^{1},L^{\infty})$ if and only if $\mathcal{F}$ is equi-integrable, that is, \begin{cases} \forall\epsilon>0\ \exists\delta>0\text{ such that}\\ \int_{A}|f|<\epsilon\ \forall A\subset\Omega,\text{measrable with }|A|<\delta,\ \forall f\in\mathcal{F} \end{cases} and \begin{cases} \forall\epsilon>0\ \exists\omega\subset\Omega,\text{ measurable with }|\omega|<\infty\text{ such that}\\ \int_{\Omega\backslash\omega}|f|<\epsilon\ \forall f\in\mathcal{F}. \end{cases} I would like to known if there is a difference between the bounded and unbounded case?

Lemma$^*$: Let $\mu_{d}$ be a (diffuse) measure in $\mathcal{M}_{d}(\Omega)$ and let $(v_{n})$ be a sequence of functions in $W^{1,p}_{0}(\Omega)$, bounded in $L^{\infty}(\Omega)$, and converging to a function $v$ $\text{cap}_{p}-$q.e., then $(v_{n})$ converges to $v$ $\mu_{d}-$a.e. and $$\underset{n\rightarrow+\infty}{\operatorname{lim}}\int_{\Omega}v_{n}d\mu_{d}=\int_{\Omega}v d\mu_{d}.$$

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    $\begingroup$ What precisely do you mean by "bounded case" and by "unbounded case"? By the way, your second sentence is not correct. $\endgroup$ Commented Jan 31, 2020 at 17:46
  • $\begingroup$ I means that $\Omega$ is a bounded/unbounded subset, this is the definition given in [H. Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, Theorem 4.30 (Dunford–Pettis)] $\endgroup$
    – M.A
    Commented Feb 1, 2020 at 18:55
  • $\begingroup$ In the ``unbounded'' case equiintegrability does not imply relative weak compactness; consider $f_n=$ the indicator function of $[n, n+1]$ in $L_1(\mathbb{R})$. $\endgroup$ Commented Feb 4, 2020 at 19:49

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