Convergence in $\mathscr{L}^1$ implies uniform integrability?

Question

Quoting from Rogers and Williams, p. 116, Theorem 21.2 (A necessary and sufficient condition for $\mathscr{L}^1$ convergence):

Let $(X_n)$ be a sequence in $\mathscr{L}^1$, and let $X\in\mathscr{L}^1$. Then $X_n\to X$ in $\mathscr{L}^1$, or, equivalently $\mathbb{E}(|X_n - X|)\to 0$, if and only if the following conditions are satisfied:

1. $X_n\to X$ in probability;
2. the sequence $(X_n)$ is uniformly integrable.

The proof provided by Rogers and Williams covers the 'if' part.

Question: But how to prove in the converse direction that convergence in $\mathscr{L}^1$ implies uniform integrability?

By the following remark in Rogers and Williams, I get the impression I should look for an integrable non-negative random variable $Y$ that bounds the collection of random variables $|X_n - X|$, but I am not sure how (e.g. $\sup_n|X_n - X|$ I suppose can't do as it might not be integrable):

"It is of course the 'if' part of the theorem that is useful. Since the result is 'best possible', it must improve on the Dominated-Convergence Theorem for our $(\Omega,\mathscr{F},\mathbb{P})$ triple; and, of course, the result (20.6) makes this explicit."

Result (20.6) reads:

Suppose that $\mathscr{C}$ is a class of random variables that is dominated by an integrable non-negative variable $Y$: $|X(\omega)|\leq Y(\omega)$ for all $X\in\mathscr{C}$ and $\mathbb{E}(Y) < \infty$. Then $\mathscr{C}$ is uniformly integrable.

Any hints? Thanks in advance!

Answer 1

Clearly, $(X_n)$ is bounded in $L^1$ so it suffices to show that $$ \forall ε>0 \ \exists δ>0 \text{ such that } (P(A)<δ \implies \mathbb E[|X_n| 1_A] \le ε \quad \text{for all } n\ge 1 ) $$ Let $ε>0$ and pick $n_0$ such that for $n> n_0$ $$ \mathbb E [|X_n-X|] <ε .$$ Note that
$$\mathbb E[|X_n| 1_A] \le \mathbb E[|X_n-X| 1_A] + \mathbb E[|X| 1_A] \le \mathbb E[|X_n-X|] + \mathbb E[|X| 1_A].$$ Hence for $n>n_0$ you have that for any measurable $A$ $$ \mathbb E[|X_n| 1_A] \le ε + \mathbb E[|X| 1_A] $$ Since $\{X_1, \dots , X_{n_0},X\}$ is uniformly integrable (this follows from the absolute continuity of the integral) we can find $δ>0$ such that $$ P(A)<δ \implies \mathbb E[|X_1|1_A] < ε, \dots , \mathbb E [|X_{n_0}|1_A]<ε \quad \text{and} \quad \mathbb E[|X|1_A]< ε . $$ In conclusion, for any measurable $A$ with $P(A)<δ$ you have that $\mathbb E[|X_n| 1_A] <2ε$ for every $n$.

Answer 2

Suppose that $(X_n)_{n\in\mathbb N}$ is not uniformly integrable. Then there exist $\varepsilon>0$ and a sequence of integers $(n_k)_{k\in\mathbb N}$ such that $$ \forall k\in\mathbb N,\quad\mathbb E\left[\vert X_{n_k}\vert1_{\{\vert X_{n_k}\vert>k\}}\right]\ge\varepsilon. $$

If $(n_k)_{k\in\mathbb N}$ were bounded, you would be able to show that the finite family $(X_n)_{0\le n\le\max_{k\in\mathbb N}n_k}$ is not uniformly integrable, which is impossible as it composed of integrable random variables.

So $(n_k)_{k\in\mathbb N}$ is not bounded, hence you can choose an increasing subsequence $(n_{k_i})_{i\in\mathbb N}$. But then you can show that $(X_{n_{k_i}})_{i\in\mathbb N}$ does not converges in $L^1$, which contradicts the convergence in $L^1$ of $(X_n)_{n\in\mathbb N}$.