I'm trying to prove the following proposition, which is another version of Dominated Convergence Theorem in Probability Theory:

Suppose $X_{n}\overset{p}\rightarrow X,$ i.e., $X_{n}\rightarrow X$ in probability, and there is a continous function $g$ with $g(x)>0$ for large $x$ with $\frac{|x|}{g(x)}\rightarrow 0$ as $|x|\rightarrow\infty$ so that $E(g(X_{n}))\leq C<\infty$ for all $n.$ Then $E(X_{n})\rightarrow E(X).$

My attempt is based in the convergence of $\frac{|x|}{g(x)}$ to $0.$ So, for an $\epsilon=1$ we have $|x|< g(x)$ for $x$ large. Then $E(|X_{n}|)<E(g(X_{n}))\leq C.$ Then $X_{n}$ are integrable and because of continuity of $g$ $g(X_{n})\rightarrow g(X)$ in probability. So $E(g(X_{n}))\rightarrow E(g(X)),$ that is because of Dominated Convergence Theorem with almost sure convergence.

But I don't know how to use this to prove the convergence of $E(X_{n})$ to $E(X).$

Any kind of help is thanked in advance.


For any $\epsilon$ and $R$, \begin{align} \left|X_n-X\right|&=\left|X_n-X\right|\mathbf 1\{\left|X_n-X\right|\leqslant \epsilon\}+ \left|X_n-X\right|\mathbf 1\{\left|X_n-X\right|\gt \epsilon\}\\ &\leqslant \epsilon+\left|X_n\right|\mathbf 1\{\left|X_n-X\right|\gt \epsilon\}+\left|X\right|\mathbf 1\{\left|X_n-X\right|\gt \epsilon\}\\ &\leqslant \epsilon+\left|X_n\right|\mathbf 1\{\left|X_n\right|\gt R\}+ R\mathbf 1\{\left|X_n-X\right|\gt \epsilon\}+\left|X\right|\mathbf 1\{\left|X\right|\gt R\} +R\mathbf 1\{\left|X_n-X\right|\gt \epsilon\}\\ \tag{*}\left|X_n-X\right|&\leqslant \epsilon+\left|X_n\right|\mathbf 1\{\left|X_n\right|\gt R\}+ 2R\mathbf 1\{\left|X_n-X\right|\gt \epsilon\}+\left|X\right|\mathbf 1\{\left|X\right|\gt R\}. \end{align} Note that $$\mathbb E\left[\left|X_n\right|\mathbf 1\{\left|X_n\right|\gt R\}\right] \leqslant \sup_{t\geqslant R}\frac t{g(t)} \mathbb E\left[g\left(X_n\right)\right]\leqslant C\sup_{t\geqslant R}\frac t{g(t)},$$ hence, integrating on both sides of (*), we get $$\mathbb E\left[\left|X_n-X\right|\right] \leqslant \epsilon+2R\cdot \mathbb P\left(\left\{\left|X_n-X\right|\gt \epsilon\right\}\right)+\mathbb E\left[\left|X\right|\mathbf 1\{\left|X\right|\gt R\}\right].$$ Using the convergence in probability of $\left(X_n\right)_{n\geqslant 1}$ to $X$,we derive that for any $R$ and $\epsilon$, $$\mathbb E\left[\left|X_n-X\right|\right] \leqslant \epsilon+\mathbb E\left[\left|X\right|\mathbf 1\{\left|X\right|\gt R\}\right].$$ To conclude, note that $\lim_{R\to +\infty}\mathbb E\left[\left|X\right|\mathbf 1\{\left|X\right|\gt R\}\right]=0$.


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