I came across this question online and am not completely sure of my answer.
On the probability space $([0,1),\mathcal{B}([0,1)),\mathbb{P}=\lambda_1)$, define the sequence of random variables $(Y_n)_{n=1}^\infty$ as follows for $(\omega \in [0,1) \space \text{and} \space n \ge 1)$
$Y_n(\omega) := \cos n\pi\omega$
Is $Y_n(\omega)$ UI (uniformly integrable)?
As I understand, if $Y_n \xrightarrow{\mathcal{L}^1} Y$ then ($Y_n$) is UI.
As I can understand we should be able to write, $$\lim_{n \to \infty} \mathbb{E} (|Y_n - Y|) = \lim_{n \to \infty} \mathbb{E} (|Y_n|) = \lim_{n \to \infty} \mathbb{E} (|\cos n\pi\omega|).$$
Now since when $n \to \infty $ the limit should not be defined since it can be anything depending on the value of $\omega$. Thus, I think $Y_n$ is not UI.
Please help me understand if this logic makes sense.
As you said, if $Y_n\to Y$ in $\mathbb L^1$, then $(Y_n)_{n\geqslant 1}$ is uniformly integrable. But it does not mean that if there is no random variable $Y$ such that $Y_n\to Y$ in $\mathbb L^1$, then $(Y_n)_{n\geqslant 1}$ is not uniformly integrable. For example, take $Y_n=(-1)^n$.
The example given in this question is an other example where $(Y_n)$ is uniformly integrable (as $\lvert Y_n(\omega)\rvert\leqslant 1$ for each $\omega$ and $n$) but without convergence in $\mathbb L^1$.