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.
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 $g$.