Prove that $X_n \to X$ in $L^1$ if and only if $E(X_n1_{A}) \to E(X1_{A})$ uniformly on $A \in \mathcal{F}$

Question

This is a probability exercise from the Karr's book called "Probability".

Prove that $X_n \overset{L^1}{\rightarrow} X$ if and only if

$$\sup_{A \in \mathcal{F}} \left|E(X_n1_{A}) - E(X1_{A})\right|\rightarrow 0.$$

Where, according to the book's notation, $X_n$ is a sequence of rv's, $A$ is an event and $\mathcal{F}$ the $\sigma$-algebra associated to $X_n$, while $1_A$ denotes the indicator function over the set $A$.

My guess is that since convergence in $L_1$ implies uniform integrability, one could use this fact to proceed, but I am completely stuck.

Answer 1

"$\Rightarrow$": Use

$$|\mathbb{E}(X_n 1_A)-\mathbb{E}(X 1_A)| \leq \mathbb{E}(|X_n-X|).$$

"$\Leftarrow$": Show

$$\mathbb{E}(|X_n-X|) = \mathbb{E}[(X_n-X) 1_{\{X_n-X \geq 0\}})]+ \mathbb{E}[(X-X_n) 1_{\{X_n-X<0\}}],$$

and conclude that

$$\mathbb{E}(|X_n-X|) \leq 2 \sup_{A \in \mathcal{F}} |\mathbb{E}[(X_n-X) 1_A]|.$$