Uniform integrability when $\forall g \in \mathcal{L}^1$ there is always an $|f_n|>|g|$ Suppose we have a sequence of measurable $f_n:X\rightarrow \mathbb{C}$ with the property that for every $g\in \mathcal{L}^1$, there is an $N$ such that $|f_N|>|g|$.
Can such a sequence be uniformly integrable?
 A: Assume that the sequence is UI and choose $\delta$ so that for all $n$ we have $\int_E f_n < 1$ for all $E$ with measure at most $\delta$. Let $A$ be any set of measure $m \leq \delta$.
Take $g_n := n 1_A$. By assumption we can find a sequence $n_k$ such that $|f_{n_k}| > k 1_A$.
But then $\int f_{n_k} >  k m > 1$ for $k$ large enough.
Thus if $X$ has sets of arbitrarily small positive measure, the sequence $f_n$ cannot be UI. 
What if $X$ does not have any sets of positive measure smaller than some $m$? Then the definition of UI says for all $\epsilon > 0$ there is $\delta > 0$ such that $\int_E f_n < \epsilon$ when $E$ has measure at most $\delta$. Well, take $\delta = m$, then any set of measure at most $\delta$ has measure zero and $\int_E f_n = 0$. On such spaces all subsets of $L^1$ are UI, including your sequence $f_n$.
A: Assume that for all integrable function $g$, there exists $N$ such that for all $x$, $\left\lvert f_n\left(x\right)\right\rvert \gt \left\lvert g\left(x\right)\right\rvert$. 
Then the sequence $\left(f_n\right)_{n\geqslant 1}$ is not even bounded in $\mathbb L^1$ hence not uniformly integrable. Indeed, the assumption implies that for each integrable function $g$, 
$$ 
\int\left\lvert g\left(x\right)\right\rvert\mathrm d\mu\left(x\right)  \leqslant \sup_{n\geqslant 1}\int\left\lvert f_n\left(x\right)\right\rvert\mathrm d\mu\left(x\right).
$$
But $\int\left\lvert g\left(x\right)\right\rvert\mathrm d\mu\left(x\right) $ can be arbitrarily large.
