Extracting a subsequence from a sequence of $\mathcal{L}^1$ functions Any help with the following problem is appreciated. 
Given: a sequence of nonnegative functions $(g_n)$ which are U.I. (uniformly integrable) in $\mathcal{L}^1(0,1)$ with $\sup_n \Vert g_n \Vert_1 < \infty$.
Claim: there exist $ g \in \mathcal{L}^1(0,1)$ and a subsequence $(g_{n_k})$ such that 
$$\forall \, \text{ measurable }E \subset [0,1]\qquad \lim_{k \to \infty} \int_E g_{n_k} =\int_E g .$$
I am having difficulty proving the claim. 
 A: Here's the sketch I would follow.


*

*Consider the measures $\nu_n$ defined by $\nu_n(E) = \int_E g_n\,dm$.  This is a bounded sequence of positive measures, so by Helly's selection theorem (or Banach-Alaoglu or something analogous) we can extract a subsequence $\nu_{n_k}$ converging vaguely to some measure $\nu$, i.e. $\int h g_n\,dm \to \int h\,d\nu$ for every $h \in C_c((0,1))$.

*Use the uniform integrability to show that $\nu$ is absolutely continuous to $m$ (the characterization of uniform integrability in terms of uniform absolute continuity would be helpful) and hence by Radon-Nikodym is of the form $\nu(E) = \int g \,dm$ for some $g$.

*For a compact set $K$, approximate $1_K$ by continuous compactly supported functions to conclude that $\int_K g_n\,dm \to \int_K g\,dm$.

*Consider the set $\mathcal{L}$ of all measurable sets $E$ for which $\int_E g_n\,dm \to \int g\,dm$.  Use a monotone class argument to show that $\mathcal{L}$ consists of all measurable sets.
If you use this for a homework problem, please credit me (and give the URL of this answer) in your submission.
A: *

*Let $\{E_j,j\in\Bbb N\}$ be an algebra generating the Borel $\sigma$-algebra of the unit interval. By boundedness, we can find a nested sequence of infinite subsets of the set of natural numbers $(I_j,j\in\Bbb N)$ such that $\left(\int_{E_j}h_nd\lambda,n\in I_j\right)$ is convergent. 

*By a diagonal argument, we can find a subsequence $(h_{n_k},k\in\Bbb N)$ such that for each $j$, the sequence $\left(\int_{E_j}h_{n_k}d\lambda,k\in \Bbb N\right)$ is Cauchy. 

*By an approximation argument and the assumption of uniform integrability, we can show that for each $E\subset [0,1]$ measurable, the sequence $\left(\int_Eh_{n_k}d\lambda,k\in\Bbb N\right)$ is convergent. 

*Let $\mu_k(E):=\int_Eh_{n_k}d\lambda$. Then $\mu_k$ is a finite measure. By 3. and this thread, the map $\mu\colon E\mapsto \lim_{k\to +\infty}\mu_k(E)$ is a measure. By boundedness in $L^1$, this is a finite measure. Furthermore, $\mu$ is absolutely continuous with respect to Lebesgue measure. Radon-Nykodym theorem gives $g$.
