Consider the following definitions(the terminology in similar questions here may differ, as well as the terminology in the general literature):
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 $(u_n)$ is equi-integrable on ${\bf R}$, but $(u_n)$ is not uniformly integrable on ${\bf R}$?
Discussion. Consider a measurable set $\Omega\subseteq {\bf R}$ such that $\lambda(\Omega)>0$ (we allow the case $\lambda(\Omega)=+\infty$).
Remark1. It is already established that, if $(u_n)$ is uniformly integrable on $\Omega$, then $(u_n)$ is equi-integrable on $\Omega$. See answers to
Reconciling uniform integrability definitions and their relation to tightness
In the question above the author of the question actually claims that "the converse is true"(this is what the question here is about), but I am not sure which assumptions on $\Omega$ and $(u_n)$ he considers.
Remark2. It is already established that, under additional assumption ${\rm sup}_{n\in{\bf N}}||u_n||_{{\rm L}^1(\Omega)}\leq C<+\infty$, we have the equivalence: $(u_n)$ is uniformly integrable on $\Omega$ iff $(u_n)$ is equi-integrable on $\Omega$. See Theorem B.104. in textbook G.~Leoni, A first course in Sobolev Spaces. 2nd ed., Graduate Studies in Mathematics 105, American Mathematical Society, Providence RI, 2017.
Remark3. It is already established that, under additional assumption $\lambda(\Omega)<+\infty$, we have the equivalence: $(u_n)$ is uniformly integrable on $\Omega$ iff $(u_n)$ is equi-integrable on $\Omega$. This follows from Theorem B.104 as well, since equi-integrability with $\lambda(\Omega)<+\infty$ implies boundedness in ${\rm L}^{1}(\Omega)$. Indeed, here is the proof:
By Theorem B.104 mentioned above, it suffices to show that, if $(u_n)$ is equi-integrable on $\Omega$, then $(u_n)$ is uniformly integrable on $\Omega$. Therefore, we assume that $(u_n)$ is equi-integrable on $\Omega$. Theorem B.104 also provides that, in order to show that $(u_n)$ is uniformly integrable on $\Omega$, it is enough to recover ${\rm sup}_{n\geq n_0}||u_n||_{{{\rm L}^{1}(\Omega)}}\leq C$. To this end, we consider $\varepsilon_0:=1$ and $\delta_0>0$ which satisfies that for every measurable $A\subseteq\Omega$ we have that $\lambda(A)\leq\delta_0$ implies ${\rm sup}_{n\geq n_0}\int_{A}|u_n|\leq\varepsilon_0$.
Since it holds that $\lambda(\Omega)<+\infty$, there exists finitely many pairwise disjoint measurable subsets $(A_k)_{k=1}^{N_0}$ in $\Omega$ such that $\Omega=\cup_{k=1}^{N_0}A_k$, and such that $\lambda(A_k)\leq\delta_0$ for every $1\leq k\leq N_0$, where $N_0=N_0(\delta_0)\in {\bf N}$. Hence, we get ${\rm sup}_{n\geq n_0}\int_{\Omega}|u_n|\leq\sum_{k=1}^{N_0}{\rm sup}_{n\geq n_0}\int_{A_k}|u_n|\leq N_0$. Q.E.D.
There is another question here with the similar proof, although the author calls "uniform integrability" what we call here "equi-integrability", so extra attention is needed when it comes to terminology:
A Question About Uniform Integrability )
Remark4. Whatever the counterexample may be (if it exists), it will satisfy that
(i) $(u_n)$ is uniformly integrable on every measurable set $\omega\subseteq {\bf R}$ such that $\lambda(\omega)<+\infty$, but $(u_n)$ is not uniformly integrable on ${\bf R}$ (because of Remark3.),
(ii) $(u_n)$ is not bounded in ${\rm L}^{1}({\bf R})$ (or even $u_n\notin {\rm L}^{1}({\bf R})$ for every $n\in {\bf N}$), because of Remark2.
(iii) $(u_n)$ is not bounded in ${\rm L}^{p}({\bf R})$ for any $p>1$ (or even $u_n\notin {\rm L}^{p}({\bf R})$ for every $n\in {\bf N}$), because of
the answer to my (closely related) question
Bounded in ${\rm L}^p({\bf R})$ for $p>1$, but not uniformly integrable on ${\bf R}$
Also, in my comments to the aforementioned question, I provided an example of the sequence $(u_n)$ which is bounded in ${\rm L}^{p}({\bf R})$ for a given $0<p<1$, but which is not equi-integrable on ${\bf R}$ (and therefore it is not uniformly integrable on ${\bf R}$).
(iv) $(u_n)$ is not bounded in ${\rm L}^{\infty}({\bf R})$ (or even $u_n\notin {\rm L}^{\infty}({\bf R})$ for every $n\in {\bf N}$), because otherwise $(u_n)$ will be uniformly integrable.
Remark5 (partial answer to my question). I was able to prove that, under the additional assumption that there exists $M_0>0$ such that
$$
(A)\;\;\; \lambda(\cup_{k=1}^{+\infty}\{|u_k|>M_0\})<+\infty\;,
$$
it holds that equi-integrability of $(u_n)$ on ${\bf R}$ implies uniformly integrability of $(u_n)$ on ${\bf R}$. Therefore, under assumption (A), we have the equivalence: $(u_n)$ is uniformly integrable on ${\bf R}$ iff $(u_n)$ is equi-integrable on ${\bf R}$. This argument works for any measurable set $\Omega\subseteq {\bf R}$ such that $\lambda(\Omega)=+\infty$ (so, not only on the domain ${\bf R}$) and measurable functions $u_n:\Omega\rightarrow {\bf R}$. The proof relies on some observations which I made in a different question(which is closed, see link below).
Proof.
We set $\Omega:={\bf R}$, $A^n_M:=\{|u_n|>M\}$, where $n\in {\bf N}$ and $M>0$. We also set $A^{\infty}_{M_0}:=\cup_{k=1}^{+\infty}A^k_{M_0}=\cup_{k=1}^{+\infty}\{|u_k|>M_0\}$. It is enough to show that it holds that $$ (B) \lim_{M\rightarrow+\infty}{\rm sup}_{n\in {\bf N}}\lambda(A^{n}_M)=0\;. $$ By the Markov-Chebyshev inequality, for $n\in {\bf N}$ and $M\geq M_0$ we have $$ (C)\;\;\;\lambda(A^n_M)\leq{{\int_{A^n_M}|u_n|}\over{M}}\leq {{\int_{A^{\infty}_{M_0}}|u_n|}\over{M}}\;, $$ where $A^{\infty}_{M_0}$ is the domain of integration which satisfies $\lambda(A^{\infty}_{M_0})<+\infty$. Since $(u_n)$ is equi-integrable on $\Omega:={\bf R}$, $(u_n)$ is also equi-integrable on $A^{\infty}_{M_0}$. Since $\lambda(A^{\infty}_{M_0})<+\infty$, by Remark 3 it follows that $(u_n)$ is bounded in ${\rm L}^{1}(A^{\infty}_{M_0})$.
(Notice that, by Remark2, it follows that $(u_n)$ is uniformly integrable on $A^{\infty}_{M_0}$. Finally, by the definition of $A^{\infty}_{M_0}$, we observe that we have the following equivalence: $(u_n)$ is uniformly integrable on $\Omega$ iff there exists $M_0>0$ such that $(u_n)$ is uniformly integrable on $A^{\infty}_{M_0}$. This already completes the proof).
In addition, to make the argument clear, we infer the following:
From the inequality (C), as we apply the ${\rm sup}_{n\in {\bf N}}$, we get $$ {\rm sup}_{n\in {\bf N}}\lambda(A^n_M)\leq {{C_0}\over{M}}\;, $$ where $M\geq M_0$ and where $C_0=C_0(M_0)<+\infty$. Hence, we get (B). Now it is easy to see that (B) implies uniform integrability of $(u_n)$ on $\Omega$. I made some remarks about this last argument in my question which is already closed, here is the link:
This closed question specifically deals with the fact that in the definition of equi-integrability we can actually allow set $E$ to depend on $n$ without changing the notion of equi-integrability.
This completes the proof (of the partial answer to my question).
Remark6. In the proof in Remark5, if we assume that there exists $M_0>0$ such that for every $n\in {\bf N}$ it holds that $$ (D)\;\;\;\lambda\{|u_n|>M_0\}<+\infty. $$ is it enough to obtain the conclusion of Remark5?. Condition (D) would be much weaker condition than (A), and easier to verify. Maybe we can introduce a version of equi-integrability for the sequence $(u_n)$ such that $u_n:\Omega_n\rightarrow {\bf R}$, where the domains $\Omega_n$ are allowed to depend on $n$. Then, possibly, we can go through the proofs of usual theorems (proofs are available in the aforementioned closed question) which explain the connection between equi-integrability and uniform integrability, and it would follow (maybe) that, under the assumption (D), we have the equivalence of equi-integrability and uniform integrability. I am not sure about this.
Remark7. I was not able to use condition (D), but here is what I was so far able to prove:
${\bf THEOREM}$.(improved partial answer to my question).
Consider an arbitrary measurable set $\Omega\subseteq {\bf R}$ (we allow the case $\lambda(\Omega)=+\infty$). Consider a sequence of measurable functions $u_n:\Omega\rightarrow {\bf R}$. Suppose that $(u_n)$ satisfies the following condition: there exists $M_0>0$ such that $$ (E)\;\;\; {\rm sup}_{n\in {\bf N}}\lambda\{s\in\Omega: |u_n(s)>M_0|\}<+\infty\;. $$
Then we have the following equivalence: $(u_n)$ is equi-integrable on $\Omega$ iff $(u_n)$ is uniformly integrable on $\Omega$.
Proof.
We only need to show that, if $(u_n)$ is equi-integrable on $\Omega$, then $(u_n)$ is uniformly integrable on $\Omega$.
We divide the proof in several steps.
Step1. We start with an adaptation of definition of equi-integrability. If $(\Omega_n)$ is a sequence of measurable sets in ${\bf R}$, and if $v_n:\Omega_n\rightarrow {\bf R}$ is a sequence of measurable functions, we say that a sequnce $(v_n)$ is an equi-integrable sequence on the sequence on the sequence of domains $(\Omega_n)$ if the following property is fulfilled:
for every $\varepsilon>0$ there exists $\delta>0$ such that for every sequence of measurable sets $(A_n)$ such that for every $n\in {\bf N}$ we have $A_n\subseteq\Omega_n$ and ${\rm sup}_{n\in {\bf N}}\lambda(A_n)\leq\delta$, it follows that ${\rm sup}_{n\in {\bf N}}\int_{A_n}|u_n|\leq\varepsilon$.
Step2. Next, we derive the analogous of Remark3. We show that, if $v_n:\Omega_n\rightarrow {\bf R}$ is an equi-integrable sequence on the sequence of domains $(\Omega_n)$, and if $$ (F)\;\;\;{\rm sup}_{n\in {\bf N}}\lambda(\Omega_n)<+\infty\;, $$ then it follows that we have $$ {\rm sup}_{n\in {\bf N}}||u_n||_{{\rm L}^1(\Omega_n)}\leq C<+\infty\;. $$ We consider $\varepsilon_0:=1$ and $\delta_0>0$ which satisfies that for every measurable $A_n\subseteq\Omega_n$ we have that ${\rm sup}_{n\in {\bf N}}\lambda(A_n)\leq\delta_0$ implies ${\rm sup}_{n\in {\bf N}}\int_{A_n}|u_n|\leq\varepsilon_0$.
Since $\lambda(\Omega_n)<+\infty$, there exists finitely many pairwise disjoint measurable subsets $(B^n_k)_{k=1}^{N_0(n)}$ in $\Omega_n$ such that $\Omega_n=\cup_{k=1}^{N_0(n)}B^n_k$, and such that ${\delta_0\over{2}}\leq\lambda(B^n_k)\leq\delta_0$ for every $1\leq k\leq N_0(n)$, where $N_0(n)=N_0(n,\delta_0)\in {\bf N}$. Hence, for every $n\in {\bf N}$ we get $$ \int_{\Omega_n}|u_n|=\sum_{k=1}^{N_0(n)}\int_{B^n_k}|u_n|\leq\sum_{k=1}^{N_0(n)}{\rm sup}_{m\in {\bf N}}\int_{B^m_k}|u_m|\leq \sum_{k=1}^{N_0(n)}\varepsilon_0=N_0(n)\varepsilon_0=N_0(n)\;. $$ Since the condition (F) holds, if $L:={\rm sup}_{n\in {\bf N}}\lambda(\Omega_n)$, then ${\rm sup}_{n\in {\bf N}}N_0(n)\leq {{2L}\over{\delta_0}}$. This completes the proof of Step2.
Step3. We set $v_n:=u_n\llcorner{\Omega^{0}_n}:\Omega^{0}_n\rightarrow {\bf R}$, where $\Omega^{0}_n:=A^n_{M_0}$. Since the condition (D) holds, it follows that the sequence of domains $(\Omega^{0}_n)$ satisfies the condition (F). Thus, by Step2, we conclude that $$ {\rm sup}_{n\in {\bf N}}||u_n||_{{\rm L}^{1}(\Omega^0_n)}\leq C\;. $$
Step 4. Finally, we continue as in Remark 5. For $M\geq M_0$ we estimate $$ {\rm sup}_{n\in {\bf N}}\lambda(|u_n|>M)\leq {\rm sup}_{n\in {\bf N}}{{\int_{\Omega^0_n}|u_n|}\over{M}}\leq {{C}\over{M}}\;, $$ getting $$ (B)\;\;\;\lim_{M\rightarrow+\infty}{\rm sup}_{n\in {\bf N}}\lambda(|u_n|>M)=0\;. $$
Q.E.D.
Conclusion. If the sequence $(u_n)$ is equi-integrable on ${\bf R}$, and if $(u_n)$ is not uniformly integrable on ${\bf R}$, then $(u_n)$ necessarily satisfies the condition which is opposite to the condition (E), namely: for every $M_0>0$ it holds that $$ {\rm sup}_{n\in {\bf N}}\lambda\{s\in {\bf R}: |u_n(s)>M_0|\}=+\infty\;. $$
