 Solution

The assertion is false. On $([0,1],\mathcal{B},\text{Lebesgue measure})$, put $X_n=n{1}_{(0,1/n)}$. Then $\mathbb{E}|X_n|=1$ for every $n$, so the family is $L^1$ bounded. For every $M$, choose $n>M$; then $\mathbb{E}[|X_n|1_{|X_n|>M}]=1$. The tail criterion for uniform integrability fails.