Weak convergence in $L^1$ implies uniform integrability.

Question

Consider the following definition of weak convergence in $L^1$.

A sequence of random variables $Z_n$ defined on $(\Omega,\mathcal{F},\mathbb{P})$ is said to converge weakly in $L^1$ to an integrable random variable $Z$ if $$ \forall E\in\mathcal{F}\Rightarrow \mathbb{E}\left[Z_n\,I(E)\right]\to\mathbb{E}\left[Z\,I(E)\right] $$

How do I prove that if $Z_n$ converge weakly in $L^1$ to $Z$, then both $Z_n$ and $Z$ are uniformly integrable? Any reference is welcome.

Answer 1

We can use the line of the answer here.

Endow $\mathcal F$ with the metric $\rho(A,B)=\mathbb P(A\Delta B)$. We can show for each $\varepsilon>0$, there is $\delta>0$ such that if $\mathbb P(A)\lt \delta$ then $|\mathbb E[Z_n\mathbf 1_A]|\lt \varepsilon$. Taking $A':=A\cap \{Z_n\leqslant 0\}$ and $A'':=A\cap \{Z_n\gt 0\}$, we can see that $\mathbb E[|Z_n|\mathbf 1_A]\lt\varepsilon$ whenever $\mathbb P(A)\lt\delta$. Indeed, for a fixed $\varepsilon\gt 0$, we define $$F_N:=\bigcap_{n\geqslant N}\left\{A\in\mathcal F,\left|\mathbb E\left[Z_n\mathbf 1_A\right]\right|\leqslant\varepsilon\right\}.$$ Each $F_N$ is closed and $\bigcup_NF_N=\mathcal F$, hence by Baire's theorem, there is $N_0$, $r_0$ and $A_0\in\mathcal F$ such that $B_\rho(A_0,r_0)\subset F_{N_0}$. Let $B$ such that $\mathbb P(B)\lt r_0$. Since $\mathbb P(A_0\Delta (A_0\cup B))\lt r_0$, $\mathbb P(A_0\Delta (A_0\cap B^c))\lt r_0$ and $$\int_B Z_n\mathrm d\mathbb P=\int_{A_0\cup B}Z_n\mathrm d\mathbb P-\int_{A_0\cap B^c}Z_n\mathrm d\mathbb P,$$ we have $\left|\int_B Z_n\mathrm d\mathbb P\right|\leqslant 2\varepsilon$ whenever $n\geqslant N_0$ and $\mathbb P(B)\lt r_0$.

Now we use Theorem 1.12.9 in Bogachev, Measure theory, volume 1:

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space. Then for each $\delta>0$, we can find an integer $N$ and a finite partition of $\Omega$, $\{S_1,\dots,S_N\}$ such that for each $i$, either $\mathbb P(S_i)\leqslant \delta$ or $S_i$ is an atom of measure $>\delta$.

So take $\varepsilon:=1$, the associated $\delta$, and notice that there are only finitely many atoms of measure $\gt \delta$. On each of these atoms, $Z_n$ is constant.

This proves that $\sup_{n\in\mathbb N}\mathbb E\left\lvert Z_n\right\rvert$ is finite. Now, in order to prove uniform integrability, we have to show that for all positive $\varepsilon$, there exists $\delta$ such that for all $n\in\mathbb N$ and all subset $A\in\mathcal F$ of probability smaller than $\delta$, $\mathbb E\left[\left\lvert Z_n\right\rvert \mathbf 1_A\right]\leqslant \varepsilon$.

We have seen that if $\mathbb P\left(A\right)\lt r_0$, then $\mathbb E\left[\left\lvert Z_n\right\rvert \mathbf 1_A\right]\leqslant 2\varepsilon$. Now, since $\max_{1\leqslant n\leqslant N-1}\left\lvert Z_n\right\rvert$ is integrable, there exists $r_1$ such that if $A\in\mathcal F$ has a probability smaller than $r_1$, $\mathbb E\left[\max_{1\leqslant n\leqslant N-1}\left\lvert Z_n\right\rvert \mathbf 1_A\right]\leqslant \varepsilon$. Therefore, we can choose $\delta:=\min\left\{r_1,r_2\right\}$.