Weak convergence in $L^1$ of a tight sequence I was reading an article on the concentration-compactness principle (P.L. Lions, concentration-compactness principle in the calculus of variations. the locally compact case, part 1, p.125) and I have to show the existence of a weak limit in $L^1$. Here is the situation:
We have a sequence of functions on $\mathbb{R}^N$, {$u_n$}, such that they are bounded in $L^1(\mathbb{R}^N) \cap L^q(\mathbb{R}^N)$. We also have that {$u_n$} is tight ($\forall \epsilon > 0$, $\exists R < \infty$, $\int_{B_R}u_n dx \geq \lambda - \epsilon$), where $u_n \geq 0$ and $\int_{\mathbb{R}^N}u_n = \lambda$.
Now (here is the part I am unsure about), taking a subsequence if necessary, we may assume that $u_n$ converges weakly to some $u$ in $L^{\alpha}(\mathbb{R}^N)$ for $1 \leq \alpha \leq q$ and $u \in L^1(\mathbb{R}^N) \cap L^q(\mathbb{R}^N)$. I agree with the fact that we can extract a weak limit in $L^{\alpha}(\mathbb{R}^N)$ for any $1 < \alpha \leq q$, passing to a diagonal subsequence  we may assume $u_n \rightharpoonup u$ in $L^{\alpha}(\mathbb{R}^N)$, $\forall 1 < \alpha \leq q$. But I am not convinced we can do it for $\alpha = 1$, since $L^1(\mathbb{R}^N)$ is not weakly closed. The only theorem in this direction I have found is the Dunford-Pettis theorem (equiintegrability guarantees weak convergence of a bounded set in $L^1$), but I am definitely not sure if this is the way to go. 
I hope the problem is clear, any help is greatly appreciated.
 A: Dufford-Pettis theorem is a promising approach. We are sure it works on probability spaces and we will use tightness assumption to be reduced to that case. 
For any integer $R$, the sequence $\left(u_n\mathbf 1_{B_R}\right)_{n\geqslant 1}$ is bounded in $\mathbb L^q$ hence if we consider the probability space $\left(B_R,\mathcal B\left(B_R\right),\lambda/\lambda\left(B_R\right)\right)$, we can extract a subsequence which converges weakly in $\mathbb L^1$ to some $u_R$, that is, such that for any measurable bounded $v$, 
$$\tag{*}    \lim_{j\to +\infty}      \int_{\mathbb R^N}u_{n_j} (x)v(x)\mathbf 1_{B_R}(x)\mathrm dx =\int_{\mathbb R^N}u_R (x)v(x)\mathbf 1_{B_R}(x)\mathrm dx.      $$
By the diagonal process, the subsequence can be chosen independently of $R$, hence (*) holds for a fixed sequence $\left(n_j\right)_{j\geqslant 1}$. 
Also, observe that $u_R=u_{R'}$ on $B_R$ if $R \lt R'$ hence we can safely remove the dependence in $R$ write $u_R=u$. For the moment, we got the following: there exists an increasing sequence of integers $\left(n_j\right)_{j\geqslant 1}$ and a function $u$ such that for any integer $R$ and any $v\in\mathbb L^\infty$, 
 $$     \lim_{j\to +\infty}      \int_{\mathbb R^N}u_{n_j} (x)v(x)\mathbf 1_{B_R}(x)\mathrm dx =\int_{\mathbb R^N}u (x)v(x)\mathbf 1_{B_R}(x)\mathrm dx.      $$
Now, we use the tightness assumption: for any positive $\varepsilon$, there exists $R$ such that for all $j$,
$$\left\lvert \int_{\mathbb R^N}u_{n_j} (x)v(x)\left(1- \mathbf 1_{B_R}(x)\right) \mathrm dx\right\rvert\leqslant \varepsilon\left\lVert v\right\rVert_{\mathbb L^\infty}              $$ 
and the same holds with $u_{n_j}$ replaced by $u$. This proves that $u_{n_j}\to u$ weakly in $\mathbb L^1$.       
