I want to show that if $X_{n}\to X$ in distribution and $X_{n}$'s are uniformly integrable then $E[|X|]<\infty$ and $E[X_{n}]\to E[X]$.
Now for the first part I argue like this.
Since the $X_{n}$'s are uniformly integrable then $E[|X_{n}|]<M$ $\forall n\in\mathbb{N}$. Where $M$ is some fixed real number. ( I have already proved that such a an M exist and is finite.)
Now by Fatou's Lemma . $E[|X|]\leq\lim\inf\,E[|X_{n}|]\leq M<\infty$.
Next I want to show that $E[X_{n}]\to E[X]$.
If I try and use Skorokhod's representation theorem(Which was only stated as a theorem in class. This is an elementary probability course). Then I can use $Y_{n}\to Y$ almost surely and $Y_{n}$ has same distributions as $X_{n}$ and $Y$ has same distribution as $X$. Now I want to use the dominated convergence theorem but I do not know what rv should I use to dominate $Y_{n}$'s .
Any help is appreciated. I do not have a background in measure theory . So please try and answer in terms of probability.
Here is a proof without appealing to Skorokhod's representation theorem:
Let $K\ge0$. The map $h_K:x\mapsto\min(K,\max(-K,x))$ is continuous and bounded (by $K$). The convergence in distribution $X_n\to X$ then implies $$\mathbb E[h_K(X_n)]\xrightarrow[n\to\infty]{}\mathbb E[h_k(X)].$$ Further, it is easy to check that $|x-h_K(x)|\le |x|\mathbf 1_{\{|x|>K\}}$ for any $x\in\mathbb R$, so by triangular inequality \begin{align*} \bigl|\mathbb E[X_n-X]\bigr|&\le\mathbb E\bigl[\lvert X_n-h_K(X_n)\rvert\bigr]+\bigl\lvert \mathbb E[h_K(X_n)-h_K(X)]\bigr\rvert+\mathbb E\bigl[\lvert h_K(X)-X\rvert\bigr]\\[.4em] &\le\mathbb E\bigl[\lvert X_n\rvert\mathbf 1_{\{\lvert X_n\rvert>K\}}\bigr]+o(1)+\mathbb E\bigl[\lvert X\rvert\mathbf 1_{\{\lvert X\rvert>K\}}\bigr]. \end{align*} This together with the uniform integrability of $(X_n)_{n\ge1}$ shows that $$\limsup_{n\to\infty}\:\bigl|\mathbb E[X_n-X]\bigr|\le\mathbb E\bigl[\lvert X\rvert\mathbf 1_{\{\lvert X\rvert>K\}}\bigr].$$ Now, letting $K\to\infty$, because $X$ is integrable (as you showed with Fatou's lemma), you can conclude by the dominated convergence theorem (or the monotone convergence theorem) that $$\lim_{n\to\infty}\mathbb E[X_n]=\mathbb E[X].$$