Pratt's lemma and convergence in distribution 
Let $(X_n)_n,$ $(Y_n)_n$ be sequences in $L^1,$ $(X_n)_n$ converges in distribution to $X \in L^1,$ $(Y_n)_n$ converges in distribution to $Y \in L^1,$ $\lim_nE[X_n]=E[X],\lim_n E[Y_n]=E[Y].$ Let $(W_n)_n$ be a sequence dominated stochastically by $(X_n)_n,$ and $(Y_n)_n$ is dominated stochastically by $(W_n)_n,$ meaning $F_{X_n}\leq F_{W_n} \leq F_{Y_n}.$ Suppose that $(W_n)_n$ converges in distribution to $W$. 
1) Prove that $W \in L^1$ and that $\lim_nE[W_n]=E[W].$ The result remains true if we only suppose that $F_{X_n}\leq F_{W_n} \leq F_{Y_n}$ on a dense set $D$?
2) If $X_n$ takes value in $[0,+\infty[$ and $Y_n$ takes value in $]-\infty,0],$ prove that, in this case, $\lim_nE[|W_n|]=E[|W|].$ 

Note that the random variables may not be defined on the same probability space.
For the first part, using $1-F_{W_n^+} \leq 1-F_{X_n^+},1-F_{W_n^-} \leq 1-F_{Y_n^-}$ and since $\lim_nF_{X_n^+}=F_{X^+} \ \lambda$-a.e$,$ $\lim_n F_{W^+_n}=F_{W^+} \ \lambda$-a.e, $\lim_nF_{W_n^-}=F_{W^-} \ \lambda$-a.e and $\lim_nF_{Y_n^-}=F_{Y^-} \ \lambda$-a.e, we conclude that $1-F_{W^+}\leq 1-F_{X^+} \ \lambda$-a.e, $ 1-F_{W^-}\leq 1-F_{Y^-} \lambda$-a.e, which means that $E[|W|]\leq E[X^+]+E[Y^-]<+\infty,$ hoping that I didn't make a mistake. 
I am stuck on this question, why $\lim_nE[W_n]=E[W]$? Also is it true that if we have the stochastic dominance on a dense set, the result remains true?
 A: $\def\d{\mathrm{d}}\def\dto{\stackrel{\mathrm{d}}{→}}\def\aeto{\xrightarrow{\mathrm{a.e}}}\def\stdom{\geqslant_{\mathrm{st}}}\def\peq{\mathrel{\phantom{=}}{}}$Even if $\{X_n\}$, $\{Y_n\}$, $\{W_n\}$ may be defined on different probability spaces, a product probability space can be constructed to embed all these random variables since only their distributions are involved in the propositions to be proved. Hence it can be assumed without loss of generality that all the random variables are in the same probability space.
For question 1, your proof of $E(|W|) < +∞$ is correct. Note that $X_n \dto X$, $Y_n \dto Y$, $W_n \dto W$ imply that $F_{X_n} \aeto F_X$, $F_{Y_n} \aeto F_Y$,
$F_{W_n} \aeto F_W$ since $F_X$, $F_Y$, $F_W$ have at most countably many points of jump. Because $X_n \stdom W_n$, so $F_{X_n} \leqslant F_{W_n}$ and\begin{align*}
&\peq E(X_n) - E(W_n)\\
&= {\small \left( \int_0^{+∞} (1 - F_{X_n}(x)) \,\d x - \int_{-∞}^0 F_{X_n}(x) \,\d x \right) - \left( \int_0^{+∞} (1 - F_{W_n}(x)) \,\d x - \int_{-∞}^0 F_{W_n}(x) \,\d x \right)}\\
&= \int_{-∞}^{+∞} (F_{W_n}(x) - F_{X_n}(x)) \,\d x.
\end{align*}
Since $X_n \dto X$ and $W_n \dto W$, then $X \stdom W$ and analogously,$$
E(X) - E(W) = \int_{-∞}^{+∞} (F_W(x) - F_X(x)) \,\d x.
$$
By Fatou's lemma,\begin{gather*}
\varlimsup_{n → ∞} (E(X_n) - E(W_n)) = \varlimsup_{n → ∞} \int_{-∞}^{+∞} (F_{W_n}(x) - F_{X_n}(x)) \,\d x\\
\leqslant \int_0^{+∞} \lim_{n → ∞} (F_{W_n}(x) - F_{X_n}(x)) \,\d x = \int_0^{+∞} (F_W(x) - F_X(x)) \,\d x = E(X) - E(W),
\end{gather*}
combining with $\lim\limits_{n → ∞} E(X_n) = E(X)$ yields $\varliminf\limits_{n → ∞} E(W_n) \geqslant E(W)$.
Because $W_n \stdom Y_n$, so $F_{W_n} \leqslant F_{Y_n}$ and analogously,$$
\varlimsup_{n → ∞} (E(W_n) - E(Y_n)) \leqslant E(W) - E(Y),
$$
combining with $\lim\limits_{n → ∞} E(Y_n) = E(Y)$ yields $\varlimsup\limits_{n → ∞} E(W_n) \leqslant E(W)$. Thus $\lim\limits_{n → ∞} E(W_n) = E(W)$.
If $F_{X_n} \leqslant F_{W_n} \leqslant F_{Y_n}$ is only known to be true on a dense set $D$, it can be proved that $F_{X_n}(x) \leqslant F_{W_n}(x) \leqslant F_{Y_n}(x)$ holds for all $x$ where $F_{X_n}$, $F_{W_n}$, $F_{Y_n}$ are continuous since for any $x \in \mathbb{R}$, there exists $\{x_n\} \subseteq D$ such that $x_n \searrow x$ when $n → ∞$ (otherwise there exists $δ > 0$ such that $D \cap (x, x + δ) = \varnothing$). Thus $X_n \stdom W_n \stdom Y_n$ is still true.
For question 2, since $X_n \stdom W_n$ implies that $X_n = X_n^+ \stdom W_n^+ \stdom 0$, so replacing $X_n$, $Y_n$, $W_n$ in question 1 by $X_n^+$, $0$, $W_n^+$ yields $\lim\limits_{n → ∞} E(W_n^+) = E(W^+)$, and analogously $\lim\limits_{n → ∞} E(W_n^-) = E(W^-)$, then$$
\lim_{n → ∞} E(|W_n|) = \lim_{n → ∞} E(W_n^+) + \lim_{n → ∞} E(W_n^-) = E(W^+) + E(W^-) = E(W).$$
A: I found a way which only uses the inequality $F_{X_n} \leq F_{W_n} \leq F_{Y_n}$ and the definition of convergence in distribution in term of continuous and bounded functions. 
$\forall n,k \in \mathbb{N},$
$$|E[W_n]-E[W]|\leq |E[W_n^+]-E[\min(W_n^+,k)]-E[W_n^-]+E[\min(W_n^-,k)]|+|E[\min(W_n^+,k)]-E[W_n^+]|+|E[W^-]-E[\min(W_n^-,k)]|$$
Since $\forall n,k \in \mathbb{N}$
$$E[W_n^+]-E[\min(W_n^+,k)]-E[W_n^-]+E[\min(W_n^-,k)]=\int_0^{+\infty}(1-F_{W_n}(x+k)-F_{W_n}(-x-k))dx$$
Also $\forall n,k \in \mathbb{N},\forall x>0,$
$$1-F_{Y_n}(x+k)-F_{Y_n}(-x-k)\leq 1-F_{W_n}(x+k)-F_{W_n}(-x-k) \leq 1-F_{X_n}(x+k)-F_{X_n}(-x-k)$$
which means that $$|E[W_n^+]-E[\min(W_n^+,k)]-E[W_n^-]+E[\min(W_n^-,k)]|\leq |E[X_n]-E[\min(X_n^+,k)]+E[\min(X_n^-,k)]|+|E[Y_n]-E[\min(Y_n^+,k)]+E[\min(Y_n^-,k)]|$$
Now we take the $\limsup_n$
$$\limsup_n|E[X_n]-E[X]|\leq |E[X]-E[\min(X^+,k)]+E[\min(X^-,k)]|+|E[Y]-E[\min(Y^+,k)]+E[\min(Y^-,k)]|+|E[\min(W^+,k)]-E[W^+]|+|E[W^-]-E[\min(W^-,k)]|$$
Taking the limit $k \to +\infty,$
$$\limsup_n |E[X_n]-E[X]|\leq 0$$
So $\lim_nE[X_n]=E[X].$
