Show that there is $f\in L^1(X,\mu)$ with $P(f)<\infty$ and $P(f_n-f)\to 0$ as $n\to\infty$

Question

Could you please help me solving this old prelim problem. Any hints are appreciated

Answer 1

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*}