If $\{f_n\}\subset L_1([0,1])$, $f_n\to f$ pointwise, and $\sup_{n} \int_{0}^{1} |f_n|\max (0, \log |f_n|)<\infty$, then $f_n\to f$ in $L_1$ I'm going through old analysis qualifying exams, and have come to a roadblock on the following problem:
Suppose that $\{f_n\}\subset L_1([0,1])$, $f_n\to f$ pointwise, and $\sup_{n} \int_{0}^{1} |f_n|\max (0, \log |f_n|)<\infty$. Show that $f\in L_1([0,1])$ and $f_n\to f$ in $L_1$.
Combining $\sup_{n} \int_{0}^{1} |f_n|\max (0, \log |f_n|)<\infty$ with Fatou's Lemma gives that $|f(x)|\max (0, \log |f(x)|) \in L_1$, and from here I can show that $f\in L_1$ by letting $A= \{x\mid |f(x)|>e\}$ and observing that $|f(x)|$ is bounded below $|f(x)|\max (0, \log |f(x)|)$ on $A$ and is bounded below $e$ on $A^c$. However, I got stuck here and am not sure how to proceed further. Hints or solutions are both welcome.
 A: Suppose, for the moment, that the family $\{f_n\}_n \cup \{f\} \subseteq L^1$ is uniformly integrable, i.e. that for any $\epsilon>0$ there exists $\delta>0$ such that
$$\sup_{n \in \mathbb{N}} \int_A |f_n| \, d\mu + \int_A |f| \, d\mu < \epsilon \tag{1}$$
for any Borel set $A$ with $\mu(A)\leq\delta$ (here, and in what follows, $\mu$ denotes a finite measure on $(\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d))$, e.g. Lebesgue measure restricted to $[0,1]$).
Fix $\epsilon>0$ and choose $\delta>0$ as above. Since $f_n \to f$ pointwise, we can choose $N \in \mathbb{N}$ sufficiently large such that $\mu(|f_n-f| \geq \epsilon) \leq \delta$ for all $n \geq N$. Using
$$\int |f_n-f| \, d\mu = \int_{\{|f_n-f| < \epsilon\}} |f_n-f| \, d\mu + \int_{\{|f_n-f| \geq \epsilon\}} |f_n-f| \, d\mu$$
we obtain from $(1)$ (with $A:=\{|f_n-f| \geq \epsilon\}$) and the triangle inequality
$$\int |f_n-f| \, d\mu \leq \epsilon \mu(X) + \int_{\{|f_n-f| \geq \epsilon\}} |f_n| \, d\mu + \int_{\{|f_n-f| \geq \epsilon\}} |f| \, d\mu \leq \epsilon(2+\mu(X))$$
for all $n \geq N$. Since $\epsilon>0$ is arbitrary, this proves $f_n \to f \in L^1$.

It remains to verify $(1)$, i.e. the uniform integrability. For given $\epsilon>0$ choose $R \geq e$ sufficiently large such that $\log(x)>\epsilon^{-1}$ for all $x \geq R$. Then
$$|f_n(x)| = \frac{1}{\log |f_n(x)|} \cdot |f_n(x) \cdot \log(|f_n(x)|)|  \leq \epsilon |f_n(x) \cdot \log(|f_n(x)|)|$$
for all $x \in \{|f_n| \geq R\}$ and therefore
$$\begin{align*} \int_{|f_n| \geq R} |f_n| \, d\mu &\leq \epsilon \int_{|f_n| \geq R} |f_n(x) \cdot \log(|f_n(x)|)) \, d\mu \\ &\leq \epsilon \sup_{n \in \mathbb{N}} \int |f_n(x) \max\{0,\log(|f_n(x)|)\}| \, d\mu=: C \epsilon \end{align*}$$
for all $R \geq R_0$. Now choose $\delta := \epsilon/R_0$, then we have for any measurable set $A$ with $\mu(A)<\delta$
$$\begin{align*} \int_A |f_n| \, d\mu &= \int_{A \cap \{|f_n| \leq R_0\}} |f_n| \, d\mu + \int_{A \cap \{|f_n| > R_0\}} |f_n| \, d\mu \\ &\leq R_0 \mu(A) + C \epsilon \leq (C+1) \epsilon. \end{align*}$$
The same reasoning works if we replace $f_n$ by $f$, and so this proves (1).
