Show that there is $f\in L^1(X,\mu)$ with $P(f)<\infty$ and $P(f_n-f)\to 0$ as $n\to\infty$ Could you please help me solving this old prelim problem. Any hints are appreciated

 A: Let $\epsilon>0$. So for $f_n\in L^1(X,\mu)$ let 
$$P(f_n)=\int_X |f_n|(1+\log^+|f_n|)\,d\mu $$
with $P(f_n)<\infty$. Let $N$ be chosen such that for all $n,m>N$ we have
$$ P(f_n-f_m)=\int_X |f_n-f_m|(1+\log^+|f_n-f_m|)\,d\mu<\epsilon.$$
Then we have
$$\int_X |f_n-f_m|\,d\mu \leq \int_X |f_n-f_m|(1+\log^+|f_n-f_m|)\,d\mu<\epsilon.$$
Hence $\{f_n\}$ is a Cauchy sequence in $L^1$ and since $L^1$ is complete, there exists a function $f\in L^1(X,\mu)$ such that $f_n\to f$ in $L^1$. Now we need to prove that $P(f)<\infty$. Let $\{f_{n_k}\}_k$ be a subsequence of $\{f_n\}$ such that $\lim_{k\to\infty} f_{n_k}(x)=f(x)$ a.e.. By Fatou's Lemma
$$\int_X |f|\,d\mu = \int_X \lim_{k\to\infty}|f_{n_k}|\,d\mu\leq \liminf_{k\to\infty}\int_X |f_{n_k}|\,d\mu<\infty.$$
Now for for all $n,n_k>N$ 
\begin{align*}
 P(f_n-f)&= \int_X |f_n-f|(1+\log^+|f_n-f|)\,d\mu\\
&= \int_X \lim_{k\to\infty}|f_n-f_{n_k}|(1+\log^+|f_n-f_{n_k}|)\,d\mu\\
&\leq \liminf_{k\to\infty} \int_X |f_n-f_{n_k}|(1+\log^+|f_n-f_{n_k}|)\,d\mu\\
&=\liminf_{k\to\infty} P(f_n-f_{n_k})\\
&<\epsilon.
\end{align*} 
