Consider a sequence $(X_n)$ of random variables satisfying $\operatorname{sup}_n E|X_n|^r< \infty$, for $r>1$. How do I show that $(X_n)$ is uniformly integrable? The answer I found is
$$E(|X_n|1_{|X_n|> A}) \leq \frac{E(|X_n|^r1_{|X_n|> A})}{A^{r-1}} \leq \frac{E(|X_n|^r)}{A^{r-1}}$$
However, I don't know how to prove the first inequality. Is there a counterexample in the case $r=1$
For the first inequality: $|X_n| = \frac{|X_n|^r}{|X_n|^{r-1}}\le \frac{|X_n|^r}{A^{r-1}}$ if $|X_n|>A$. To deduce the second inequality: $|X_n|^r1_{|X_n|>A}\le |X_n|^r$ holds everywhere, because if $|X_n(x)|>A$, then the two sides are equal, and otherwise, the LHS $=0$.
In case of $r=1$, we can take e.g. $x$ be a uniform random variable over $[0, 1]$, and $X_n = 2^n\cdot1_{[0, 2^{-n}]}$, then $E|X_n| \equiv 1$ is uniformly bounded, but clearly not uniformly integrable.