 Solution

For a fixed $c>0$, the functions $|f_n|1_{\{|f_n|\le c\}}$ tend pointwise to zero and are dominated by the Lebesgue integrable constant $c$, since $\mu (\Omega )<\infty$. The dominated convergence theorem therefore gives convergence of their integrals to zero. SplitTake the upper limit in $n$ and then send $c\to \infty$. The specified uniform integrability condition makes the right side zero. It also ensures integrability of each $f_n$, by the same split for a sufficiently large $c$. Hence $\int |f_n|d\mu \to 0$, and in particular $\int f_{n}d\mu \to 0$.

For a counterexample without the tail condition, use Lebesgue measure on $(0,1)$ and $f_n(x)=n{1}_{(0,1/n)}(x)$. Every fixed positive $x$ eventually lies outside the shrinking interval, so $f_n(x)\to 0$, but ${\int }_0^1{f}_{n}dx=1$ for every $n$.