Consider the following definitions.
Definition.(UNIFORM INTEGRABILITY). Consider a measurable set $\Omega\subseteq {\bf R}$ such that $\lambda(\Omega)>0$ and a sequence of measurable functions $u_n:\Omega\rightarrow {\bf R}$. We say that a sequence $(u_n)$ is uniformly integrable on $\Omega$ if $\lim_{M\rightarrow+\infty}{\rm sup}_{n\in{\bf N}}\int_{\{|u_n|>M\}}|u_n|=0$ (also known as the condition of "NO ESCAPE TO VERTICAL INFINITY").
Definition.(EQUI-INTEGRABILITY). Consider a measurable set $\Omega\subseteq {\bf R}$ such that $\lambda(\Omega)>0$ and a sequence of measurable functions $u_n:\Omega\rightarrow {\bf R}$. We say that a sequence $(u_n)$ is equi-integrable on $\Omega$ if for every $\varepsilon>0$ there exists $\delta>0$ such that for every measurable set $E\subseteq\Omega$ such that $\lambda(E)\leq\delta$ it follows that for every $n\in {\bf N}$ we have $\int_{E}|u_n|\leq\varepsilon$.
Question: Is it possible to construct a sequence of measurable functions $u_n:{\bf R}\rightarrow {\bf R}$ such that
A (weak version). Let $p>1$ be given.
(A.I) ${\rm sup}_{n\in {\bf N}}||u_n||_{{\rm L}^p({\bf R})}\leq C$ (so $(u_n)$ is bounded in ${\rm L}^p({\bf R})$),
(A.II) $\lim_{M\rightarrow+\infty}{\rm sup}_{n\in {\bf N}}\int_{\{|u_n|>M\}}|u_n|>0$ (so $(u_n)$ is not uniformly integrable, i.e. $(u_n)$ exhibits "escape to vertical infinity").
B (strong version).
(B.I) for every $1<p<+\infty$ we have ${\rm sup}_{n\in {\bf N}}||u_n||_{{\rm L}^p({\bf R})}\leq C_p<+\infty$ (so $(u_n)$ is bounded in ${\rm L}^p({\bf R})$ for every $1<p<+\infty$),
(B.II) $\lim_{M\rightarrow+\infty}{\rm sup}_{n\in {\bf N}}\int_{\{|u_n|>M\}}|u_n|=+\infty$ (so $(u_n)$ is not uniformly integrable, i.e. $(u_n)$ exhibits strong "escape to vertical infinity").
Discussion.
By Theorem B.104 in G. Leoni, A first course in Sobolev Spaces. 2nd ed.}, Graduate Studies in Mathematics 105, American Mathematical Society, Providence RI, 2017. (also cf. Theorem 2.29 in I. Fonseca and G. Leoni, Modern Methods in the Calculus of Variations: ${\rm L}^p$-spaces}, Studies in Advanced Mathematics, Springer, New York 2007.), under the assumption of boundedness in ${\rm L}^1({\bf R})$, equi-integrability is equivalent to uniform integrability.
If $\lambda(\Omega)<+\infty$, then condition A.(I) (by Holder's Inequality) implies that $(u_n)$ is equi-integrable on $\Omega$, and therefore $(u_n)$ is bounded in ${\rm L}^1(\Omega)$, see
A Question About Uniform Integrability
As a consequence, it is not possible to construct an example which satisfies both A.(I) and A.(II) (both B.(I) and B.(II), resp.) on domains $\Omega\subseteq {\bf R}$ of finite measure. See also:
uniform integrability does not imply $L_p$ boundednes
Compare also questions
Family of functions that are bounded in $L^1$ but *NOT* Uniformly Integrable
Remark1. In the cited questions above, the notions of uniform integrability are defined in different ways, which can be confusing.
Remark2. If $(u_n)$ satisfies the condition $\lim_{M\rightarrow+\infty}{\rm sup}_{n\in {\bf N}}\lambda(|u_n|>M)=0$, then, by Holder's Inequality, we estimate $\int_{\{|u_n|>M\}}|u_n|\leq \Big(\int_{\{|u_n|>M\}}|u_n|^p\Big)^{1\over{p}}\lambda(|u_n|>M)^{1\over{p'}}$, where ${1\over{p}}+{1\over{p'}}=1$, and it follows that condition A.(I) implies $\lim_{M\rightarrow+\infty}{\rm sup}_{n\in {\bf N}}\int_{\{|u_n|>M\}}|u_n|=0$, so A.(II) can not be satisfied.
Remark3. If $(u_n)$ satisfies the condition $\lim_{M\rightarrow+\infty}{\rm sup}_{n\in {\bf N}}\lambda(|u_n|>M)<+\infty$, then, by Holder's Inequality, we estimate $\int_{\{|u_n|>M\}}|u_n|\leq \Big(\int_{\{|u_n|>M\}}|u_n|^p\Big)^{1\over{p}}\lambda(|u_n|>M)^{1\over{p'}}$, where ${1\over{p}}+{1\over{p'}}=1$, and it follows that condition B.(I) implies $\lim_{M\rightarrow+\infty}{\rm sup}_{n\in {\bf N}}\int_{\{|u_n|>M\}}|u_n|<+\infty$, so B.(II) can not be satisfied.
Conclusion: B.(II) can not be satisfied, even under assumption (A.I), which is weaker then the assumption B.(I). To see that this is the case, for $M\geq 1$ and for a given $p>1$ we estimate $$ {\rm sup}_{n\in {\bf N}}\lambda(|u_n|>M)={\rm sup}_{n\in {\bf N}}\int_{|u_n|>M}1\leq {\rm sup}_{n\in {\bf N}}\int_{|u_n|>M}|u_n|\leq {\rm sup}_{n\in {\bf N}}\int_{|u_n|>M}|u_n|^p\;. $$ Hence, under assumption (A.I), it holds that $\lim_{M\rightarrow+\infty}{\rm sup}_{n\in {\bf N}}\lambda(|u_n|>M)<+\infty$, and therefore (B.II). can not be satisfied. So the strong version can not be constructed. On the other hand, to satisfy the conditions of the weak version, it is necessary that $(u_n)$ satisfies the condition $$\lim_{M\rightarrow+\infty}{\rm sup}_{n\in {\bf N}}\lambda(|u_n|>M)>0\;.$$
Remark4. If $p>0$, the sequence $u_n(s):=\sum_{k=1}^{n}2^k\chi_{(2^k,2^k+2^{-np})}(s)$, where $s\in {\bf R}$, satisfies: $$ ||u_n||_{{\rm L}^p({\bf R})}^p\leq \Big(2^{p}-1\Big)+\sum_{j=0}^{+\infty}(2^{-p})^j\;, $$
$$ ||u_n||_{{\rm L}^1({\bf R})}=2^{-np}\Big(2^{n+1}-1\Big)\;, $$
$$ \int_{|u_n|>M}|u_n|\geq\int_{\cup_{i=M_0}^{n}\{\sigma\in (2^{i}, 2^{i}+2^{-np}):|u_n(\sigma)|>M\}}|u_n|=\sum_{i=M_0}^{n}\int_{2^i}^{2^i+2^{-np}}2^i=2^{n(1-p)+1}-2^{M_0-np}\;, $$
where $M>2$ and $M_0>{\rm log}_{2}M>1$. Therefore, $(u_n)$ provides the example of a sequence which is bounded in ${\rm L}^p({\bf R})$ for $0<p<1$, but which is not uniformly integrable on ${\bf R}$.
