Let $(X,\mathcal{M},\mu)$ be a measure space. Show that if $f: X \rightarrow \mathbb{C}$ is integrable, then there exists $f_n$ which are integrable and $f_n \rightarrow f$ pointwise and in $L^1$.
Attempt: The question says not to use the fact that $f_n \rightarrow f$ in $L^1$ implies there is a subsequence $f_{n_j} \rightarrow f$ pointwise. However, my method relies on this notion. Is there any different way to do this?
Since $f$ is integrable, for each $n \in \mathbb{N}$, there is a continuous function $g_n \in L^1$ vanishing outside a bounded interval s.t. $\int|g_n-f|<\frac{1}{n}$. Thus, $\int|g_n-f| \rightarrow 0$ and $g_n \rightarrow f$ in $L^1$. Since $g_n \rightarrow f$ in $L^1$, there is a subsequence such that $g_{n_j} \rightarrow f$ pointwise. Further, $g_{n_j} \rightarrow f$ in $L^1$.
Edit: I realize my approach is badly flawed, so I will try to use useful comments to do this. Let $f_n=(1-\frac{1}{n})f$. Then $f_n \rightarrow f$ pointwise since $1-\frac{1}{n} \rightarrow 1$. Since $|(1-\frac{1}{n})f|\leq |f|$, by the dominating convergence theorem, $f_n \rightarrow f$ in $L^1$.