# Convergence in distribution implies convergence of $L^p$ norms under additional assumption

The following is an exercise (4.5.2) from Chung's book, "A Course in Probability Theory".

If $\{X_n\}$ is dominated by some $Z$ in $L^p$ and converges in distribution to $X$, then prove that $$\lim\limits_{n\rightarrow\infty}E[|X_n|^p]=E[|X|^p]$$

I applied the Skorohod's Representation theorem. Since $X_n$ converges in distribution to $X$, there exists a sequence of random variables $\{Y_n\}$ with the same distribution as $\{X_n\}$ such that $Y_n \rightarrow Y$ a.s and $Y$ has the same distribution as $X$. By the continuous mapping theorem,

$$\lim\limits_{n\rightarrow\infty}|Y_n|^p = |Y|^p$$

almost surely. Then,

$$\lim\limits_{n\rightarrow\infty}E[|X_n|^p] = \lim\limits_{n\rightarrow\infty}E[|Y_n|^p] = E[|Y|^p] = E[|X|^p]$$

where I applied the Dominated Convergence Theorem in the second step. Am I allowed to apply the Dominated Convergence Theorem here? We know that $\{X_n\}$ are dominated by $Z$, but does that mean the $\{Y_n\}$ are as well?

No, your argument is not correct because $\{Y_n\}$ need not be dominated in $L^{p}$. Note that $$\int_{|X_n|^{p}>\Delta } |X_n|^{p} \leq \int_{Z^{p}>\Delta } Z^{p} \to 0$$ uniformly in $n$ as $\Delta \to \infty$. Now use Skhrohod Representation and get $\int_{|Y_n|^{p}>\Delta } |Y_n|^{p} \to 0$ uniformly in $n$ as $\Delta \to \infty$. Since $Y_n-Y \to 0$ almost surely and $\{|Y_n-Y|^{p}\}$ is uniformly integrable we get $E|Y_n-Y|^{p} \to 0$. From this we get $$\limsup E|Y_n|^{p} \leq E|Y|^{p}$$ so $$\limsup E|X_n|^{p} \leq E|X|^{p}$$. On the other hand $$\liminf E|Y_n|^{p} \geq E|Y|^{p}$$ by Fatou's Lemma which gives $$\liminf E|X_n|^{p} \geq E|X|^{p}$$.