convergence in measure implies integral goes to zero Prove if $f_n$ converges to $0$ in measure and $\mu$ is a finite measure, then
$$\lim_{n \rightarrow \infty} \int_X \frac{f_n}{1+\vert f_n \vert} d \mu = 0$$
So I know that if $f_n$ converges to $0$ in measure, then
$$\mu(\{x \in X: \vert f_n(x) \vert > \epsilon\})=0$$
Any hints would be greatly appreciated. I wanna say I can maybe bound the integral using triangle inequality? But unsure if that would work.

Hint: Let $E_n = \{x \in X: |f_n(x)| > \epsilon\}$ and note that
$$x \in E_n \implies\frac{|f_n|}{1+|f_n|} > \frac{\epsilon}{1 + \epsilon}, \quad x \in X \setminus E_n  \implies\frac{|f_n|}{1+|f_n|} \leqslant \frac{\epsilon}{1 + \epsilon}$$
