Weak Convergence in the Space $L_1$, Why So Special? For $1< p< \infty$ and $U\subset \mathbb{R}^d$, and let $p'$ be the conjugate of $p$. We say that a sequence $\{f_n\}\in L^p(U)$ converges weakly to $f\in L^p(U)$ if 
$$ \lim_{n\to \infty}\int_U f_n(u) g(u) d(u)=\int_U f(u)g(u) d(u) \ \ \ \  \forall g\in L^{p'}(U).  $$
Knowing a small amount of probability I think of weak-convergence as convergence in average (mentioned in these notes https://www.uio.no/studier/emner/matnat/math/MAT4380/v06/Weakconvergence.pdf Remark 1.2). 
My question is what is so special about $L^1-$convergence? What are the usual ways to guarantee a sequence converges weakly somewhere in $L^1$ ?
 A: First, a few general notions: Weak convergence of a sequence $(u_n)$ to $u$ in a Banach space $X$ means that $f(u_n)\rightarrow f(u)$ for all $f\in X^*.$ Weak* convergence of a sequence $(f_n)$ to $f$ in $X^*$ means that $f_n(u)\rightarrow f(u)$ for all $u\in X$. 
Recall that for $1\leq p<\infty$ (and a nice enough measure space), $L^{p'}$ is isometrically isomorphic to the dual of $L^p$, where $1/p+1/p'=1,$ via the map $v\rightarrow I_v$ given by $$I_v(u)=\int uv\, dx,$$ where  $u\in L^p.$ So, weak and weak* convergence get recast in this way.
For $1<p<\infty,$ the $L^p$-spaces are reflexive, so the notions of weak and weak* are equivalent. While $L^1$ is not reflexive, it is true that the dual of $L^1(U)$ is $L^\infty(U)$, so weak* convergence of $(u_n)$ to $u$ in $L^\infty(U)$ means that $$\int_U u_n v\, dx\rightarrow \int_U uv\, dx$$ for all $v\in L^1(U).$ This is why the link makes a distinction when $p=\infty.$ Similarly, $(u_n)$ converges weakly to $u$ in $L^1(U)$ provided $$\int\limits_U u_nv\, dx\rightarrow \int_U uv\, dx$$ for all $v\in L^\infty(U).$
