Let $(X,\Omega,\mu)$ be a measure space. A sequence $f_n$ is said to be uniformly integrable if for every $\epsilon \gt 0$ there is a $\delta \gt 0$ such that for every measurable set $A$ with $\mu(A)\lt \delta$, $\int_A |f_n|~d\mu \lt \epsilon$, for every $n\in \mathbb{N}$.
A sequence is said to be tight if for every $\epsilon \gt 0$ there is a measurable set $B$ of finite measure such that $\int_{X\setminus B} |f_n|~d\mu \lt \epsilon $, for every $n\in \mathbb{N}$.
I claim the $f_n = n\cdot 1_{[0,1/n]}$ is not uniformly integrable and $g_n = 1_{[n,n+1]}$ is not tight.
Proof. Fix $\epsilon \gt 0$. Pick $n$ sufficiently large so that for every $\delta \gt 0$, $n\delta \gt 1/2.$ Then there is an $n$ such that $\int_A |f_n| \gt 1/2.$
Let $\mu(B)\lt \infty$. Suppose to the contrary that $g_n$ were tight. Then $\mu\left((X\setminus B)\cap [n,n+1]\right) \lt \epsilon$. If I can get that $\mu(B) = \infty$, then I would have a contradiction, but I don't see how.
Is what I have done above right?