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:
- $X_n\to X$ in probability;
- 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!
Best Answer
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 $n$.