Uniform integrability when $\forall g \in \mathcal{L}^1$ there is always an $|f_n|>|g|$

Question

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?

Answer 1

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 $f_n$.

Answer 2

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.