convergence of truncated sequence in $L_{1}$

Question

Consider $(f_n) \subset L_{1}(X,\mu)$, where $(X,\mu)$ is a finite measure space. Further assume that $f_n$ converges in norm to $f\in L_{1}(X,\mu)$. Now consider the truncated elements $g_n:=\chi_{ \{|f_n|>C \}}\cdot f_n$ and $C>0$. Does it follow that $g_n$ converges to $g:=\chi_{ \{|f|>C \}}\cdot f$?

Answer 1

Counter-example: $f_n=1+\frac 1 n, f=1, C=1$ on $(0,1)$ with Lebesgue measure.

Answer 2

In general no: take $f_n=C+\mathbf{1}_{(0,1/n)}$, where $X$ is the unit interval. Then $g_n=\mathbf{1}_{[1/n,1)}\left(C+\mathbf{1}_{(0,1/n)}\right)=C\mathbf{1}_{[1/n,1)}$ while $f=C$ (hence $g=0$).