# Uniform integrability and weak L1 convergence

I am working on exercise 4.14 in chapter 3 (on convergence) in the book "Probability and Stochastics" by Erhan Cinlar. The exercise can be found on page 109.

First, let me give the necessary definitions to make sure the exercise is understood correctly:

A sequence $(X_n)$ in $L^1$ is said to converge weakly in $L^1$ if the following holds (p. 108): $$\lim \mathbb{E} X_n Y = \mathbb{E} X Y,$$ for every bounded random variable $Y$.

A sequence $(X_n)$ is uniformly integrable (p. 72) if:$$\limsup \mathbb{E}[|X_n|1_{\{|X_n|>b\}}] =0.$$

Selection principle (p. 95): If every subsequence that has a limit has the same value for the limit, then the sequence tends to the saem value. If the sequence is bounded, and every convergent subsequence of it has the same limit, then the sequence converges to the same value.

Finally, here is the exercise: A sequence is uniformly integrable if and only if its every subsequence has a further subsequence that converges weakly in $L^1$.

Here is my current approach (which I think is wrong):

$\Rightarrow:$ Uniform integrability is equal to convergence in $L^1$, so if there is a bounded random variable Y (i.e. $|Y| < b$), then $$|\mathbb{E}X_n Y - \mathbb{E}XY| \leq \mathbb{E}|X_n Y - XY| \leq b \mathbb{E}|X_n - X| \rightarrow 0.$$

$\Leftarrow:$ We know there exists a bounded random variable Y such that $\lim \mathbb{E}X_n Y = \mathbb{E} XY,$so we can use the selection principle to show that $\lim_{j \rightarrow \infty} \mathbb{E}[X_{n_j}Y] = E[XY].$ Applying the selection principle again, we can show that $\lim_{n \rightarrow \infty}\mathbb{E}[X_nY] =\mathbb{E}[XY].$ Therefire $X_nY$ converges to XY in $L^1$, and therefore it is uniformly integrable.