$\quad$ Let $E\subset \mathbb R^d$ is a measurable set, on which there is an integrable function sequence $\{f_n\}$,i.e. $f_n\in\mathcal L^1(E),n\ge1$, which satisfies $f_n\to f,\mathrm{a.e.}$ and satisfies the following "integral finite condition": \begin{equation} \sup_n\int_E|f_n|\ln(2+|f_n|)<\infty. \end{equation} $\quad$ Question: Does $f_n$ converge in $\mathcal L^1$? a.e. \begin{equation} \lim_{n\to\infty}\int_E|f_n-f|=\lim_{n\to\infty}\left\| f_n-f\right\|_{\mathcal L^1}=0. \end{equation}
Edit: Thanks to @Cauchy's Sequence for the counterexample $\frac1n\chi_{[0,n]}$. What happens to this problem if we add the condition that the measure of $E$ is finite: $m(E)<\infty$? Can we give another counterexample that satisfies this condition? Or now can this proposition proved to be true?
Let $E = \mathbb{R}$ and $f_n = \frac{1}{n} \chi_{[0, n]}$ for $n \ge 1$. Clearly $|f_n| \le 1$ for all $n$ and $f_n \to 0$ a.e. So all of your hypotheses are satisfied but $f_n$ does not converge in $L^1$ since $$ \int_{\mathbb{R}} |f_n| = 1 $$ for all $n$.
The story is different if $E$ has finite measure. The answer is positive in this case. To see this, you need the following special case of the Vitali convergence theorem: Suppose $(f_n)$ is a uniformly integrable sequence in $L^1(E, \mu)$, i.e.
$$ \lim_{M \to \infty} \sup_n \int_{|f_n| \ge M} |f_n| \, d\mu = 0. $$
If $(f_n)$ converges in measure to $f$, then it also converges in $L^1(E, \mu)$ to $f$.
Now, set $\Phi(x) = x \ln(2 + x)$. Clearly, $\frac{\Phi(x)}{x} \to \infty$ as $x \to \infty$. Using this property and the hypothesis
$$ C := \sup_n \int_E \Phi(|f_n|) \, d\mu < \infty, $$
we show that $(f_n)$ is uniformly integrable. Choose $N$ large enough so that $\frac{C}{N} < \varepsilon$. There exists $M_0$ such that $\Phi(x) > Nx$ for all $|x| \ge M_0$. If $M \ge M_0$, then
$$ \int_{|f_n| \ge M} |f_n| \, d\mu \le \int_{|f_n| \ge M} \frac{\Phi(|f_n|)}{N} \, d\mu \le \frac{C}{N} < \varepsilon$$
for all $n \in \mathbb{N}$. Thus, $(f_n)$ is uniformly integrable.
The claim then follows from Vitali's convergence theorem.