I assume $f$ integrable, otherwise it's not necessarily true, as copper.hat shows.
Fix an integer $k$, and $E_k$ measurable such that $\mu(X\setminus E_k)\leq k^{-1}$ and $\sup_{x\in E_k}|f_n(x)-f(x)|\to 0$. We have, using the hypothesis, \begin{align} \int_X|f-f_n|d\mu&=\int_{E_k}|f-f_n|d\mu+\int_{X\setminus E_k}|f-f_n|d\mu\\ &\leq \mu(E_k)\sup_{x\in E_k}|f(x)-f_n(x)| +\int_{X\setminus E_k}|f-f_n|d\mu\\ &\leq \mu(X)\sup_{x\in E_k}|f(x)-f_n(x)| +\int_{X\setminus E_k}|f-f_n|d\mu. \end{align} Taking $\limsup_{n\to +\infty}$, we get $\limsup_{n\to +\infty}\int_X|f-f_n|d\mu\leq \limsup_{n\to +\infty}\int_{X\setminus E_k}\left(|f-f_n|-(|f_n|-|f|)\right)d\mu,$ using the fact that $\int_{X\setminus E_k}f_nd\mu\to \int_{X\setminus E_k}fd\mu$.
As $0\leq |f-f_n|-(|f_n|-|f|)\leq 2f$, we have for each $k$, $\limsup_{n\to +\infty}\int_X|f-f_n|d\mu\leq \int_{X\setminus E_k}fd\mu.$ Let $s_l$ a simple function such that $0\leq s_l\leq f$ and $\int_X (f-s_l)d\mu\leq l^{—1}$. We can write $s_l(x)=\sum_{j=1}^{N_l}a_{j,l}\chi_{B_{j,l}}$, where $B_{j,l}$ are measurable sets and $a_{j,l}$ positive real numbers. Then for each $k,l$ positive integers, \begin{align} \limsup_{n\to +\infty}\int_X|f-f_n|d\mu&\leq\int_{X\setminus E_k}(f-s_l)d\mu+ \int_{X\setminus E_k}s_ld\mu\\ &\leq \int_X(f-s_l)d\mu+\sum_{j=1}^{N_l}a_{j,l}\mu(B_{j,l}\cap E_k^c)\\ &\leq l^{-1}+\sum_{j=1}^{N_l}a_{j,l}\mu(B_{j,l}\cap E_k^c)\\ &\leq l^{-1}+k^{-1}\sum_{j=1}^{N_l}a_{j,l}\mu(B_{j,l}). \end{align} Taking the $\limsup_{k\to +\infty}$ then $\limsup_{l\to +\infty}$, we get the result.