= Solution
For a fixed $c>0$, the functions $|f_n|\mathbf1_{\{|f_n|\le c\}}$ 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. Split
$$
\int|f_n|d\mu\le
\int |f_n|\mathbf1_{\{|f_n|\le c\}}d\mu+
\sup_k\int_{|f_k|>c}|f_k|d\mu.
$$
Take 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 \b[$\int|f_n|d\mu\to0$], and in particular $\int f_n d\mu\to0$.
For a counterexample without the tail condition, use Lebesgue measure on $(0,1)$ and $f_n(x)=n\mathbf1_{(0,1/n)}(x)$. Every fixed positive $x$ eventually lies outside the shrinking interval, so $f_n(x)\to0$, but $\int_0^1f_n dx=1$ for every $n$.
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