Let $\{X_n\}$ be a sequence of random variables such that the sequence $\{\sqrt{n}(X_n+o_p(X_n))\}$ is uniformly integrable. Does it imply that $\{\sqrt{n}X_n\}$ is also uniformly integrable?
Note: $o_p(X_n)$ means that it is a sequence of random variable $Y_n$ such that $\frac{Y_n}{X_n}\overset{P}{\to}0$.
Replacing $X_n$ by $X'_n:=\sqrt nX_n$, the question can be rephrased as follows: if $\left(X_n\right)_{n\geqslant 1}$ is a sequence of random variables such that $\{X_n(1+\varepsilon_n),n\geqslant 1\}$ is uniformly integrable for a sequence $(\varepsilon_n)_n$ converging to $0$ in probability, is $\{X_n,n\geqslant 1\}$ uniformly integrable?
The answer is: in general no. Let $X_n$ be a random variable taking the value $n$ with probability $1/n$ and the value $0$ with probability $1-1/n$. The sequence $\{X_n(1+\varepsilon_n),n\geqslant 1\}$ is not uniformly integrable. Now take $\varepsilon_n$ as the random variable defined by $-\mathbf 1\{X_n=n\}$, where $\mathbf 1$ denotes the indicator function. Then $ X_n(1+\varepsilon_n)=0$ and $\varepsilon_n\to 0$ in probability since the random variable $\varepsilon_n$ takes a non-zero value with probability $1/n$.