$L^1$ convergence and uniform convergence of $\mathbb E X_n 1_H$. 
Convergence of expectations. Let $X_1, X_2, ..., X$ be in $L^1$. Show that $X_n \to X$ in $L^1$ if and only if $\mathbb E X_n 1_H \to \mathbb E X 1_H$ uniformly in $H$ in $\mathcal H$, that is, if and only if
\begin{equation}
\lim_n \sup_{H \in \mathcal H}|\mathbb EX_n 1_H - \mathbb E X 1_H| = 0.
\end{equation}

For necessity, I can see that $|\mathbb EX_n1_H - \mathbb E X 1_H| \le \mathbb E|X_n-X|$ for all $H \in \mathcal H$ and $n$. For sufficiency, my attempt is as follows. First, $\mathbb E|X_n - X| \le \mathbb E|X_n-X|1_{H_n} + \epsilon$ for $H_n = \{|X_n-X| > \epsilon\}$. I wish the first term goes to zero as $n \to \infty$. But, $|\mathbb E(X_n-X)1_{H_n}| = \mathbb E|X_n-X|1_{H_n}$ along $N$ where $N$ is a subsequence such that $X_n \ge X$ for $n \in N$. It doesn't look like a right way to solve this. Can you give some hint for a sufficiency part?
 A: Our objective:
For sufficiency, you need to prove that:
$\lim_{n\to\infty} \mathbb{E}[|X_n-X|] = 0$
and we know that:
$\lim_{n\to\infty} \sup_{H\in\mathcal{H}}\{|\mathbb{E}[(X_n-X)\mathbb{1}_H]|\} = 0$
Brief motivation
The only thing in which both really differ is where find the modulus function; if we had for example $X_n\leq X$ almost surely for every $n\in\mathbb{N}$ then it would be substantially easier to prove the result (try it!).
A solution: as $|X_n-X| = X-X_n\geq 0$ almost surely then $\mathbb{E}[(X-X_n)\mathbb{1}_H]\geq 0$ for every $H\in\mathcal{H}$ and then $\sup_{H\in\mathcal{H}}\{|\mathbb{E}[(X_n-X)\mathbb{1}_H]|\} = \mathbb{E}[X-X_n] = \mathbb{E}[|X-X_n|]$ and as $n$ goes to infinity we know this goes to $0$.
How can we adapt it for the general setting?
If we knew then that the random variables $X_n-X$ were almost surely positive or almost surely negative for each $n\in\mathbb{N}$, then similar arguments would allow us to prove the result. However, that need not be the case, and there can be regions in which $X_n-X$ is positive and $X_n-X$ is negative, which may vary for differnt $n\in\mathbb{N}$. However, we can control the expectation of $X_n-X$ over regions of our probability space, so our challenge will be choosing adequate regions so that the random variables $X_n-X$ become simpler, similar to the almost surely positive or almost surely negative case in which we know we can prove the result (which regions do you think will be adequate?). To reiterate, if we can replace the complicated $|X_n-X|$ by just $X_n-X$ in regions we know will be reasonable then you may just be able to adapt the solution for the simple case.
One final hint:
One of these two facts may inspire you:
$$|X_n-X|(\omega) = (X_n-X)^{+}(\omega) + (X_n-X)^{-}(\omega)$$
and
$\Omega = \{\omega \in \Omega | X_n(\omega)\geq X(\omega)\} \cup \{\omega \in \Omega | X_n(\omega)< X(\omega)\}$ where the union is disjoint.
Solution:
We try to replace the modulus by finite sums of $X_n-X$ multiplied by indicator functions, since those are the expectations we know we can control.
\begin{align*}
\mathbb{E}[|X_n-X|] 
&= \mathbb{E}[(X_n-X)1_{\{\omega \in \Omega | X_n(\omega)\geq X(\omega)\}} - (X_n-X)1_{\{\omega \in \Omega | X_n(\omega)< X(\omega)\}}] \\
&= \mathbb{E}[(X_n-X)1_{\{\omega \in \Omega | X_n(\omega)\geq X(\omega)\}}] - \mathbb{E}[(X_n-X)1_{\{\omega \in \Omega | X_n(\omega)< X(\omega)\}}] \\
&= \mathbb{E}[(X_n-X)1_{\{\omega \in \Omega | X_n(\omega)\geq X(\omega)\}}] + \mathbb{E}[(X-X_n)1_{\{\omega \in \Omega | X_n(\omega)< X(\omega)\}}] \\
&\leq 2\cdot \sup_{H\in\mathcal{H}}\{|\mathbb{E}[(X_n-X)\mathbb{1}_H]|\}
\end{align*}
from which we can then conclude that the desired limit as $n$ goes to infinity is $0$.
