Convergence in metric and in measure

Let $\mu$ be a finite measure on $(X, A)$, with the semimetric $$ d(f,g) = \int \frac{|f-g|}{1+ |f-g|}d\mu$$ on all real-valued, A-measurable functions. Show that $$\lim_n d(f_n, f) = 0$$ holds iff $(f_n)$ converges to $f$ in measure.

I know that convergence in mean implies convergence in measure but $ \int \frac{|f-g|}{1+ |f-g|}d\mu \leq \int {|f-g|}d\mu $ and also $ \mu(\{x\in X\ : |f_n(x) - f(x) > \epsilon\}) \leq \int {|f-g|}d\mu $. So I don't know what to do.

First, a remark: the map $x\mapsto \frac x{1+x}$ is increasing over the set of non-negative real numbers, and is bounded by $1$.

If $f_n\to f$ in measure, fix $\varepsilon$ and integrate over $\{|f_n-f|>\varepsilon\}$ and $\{|f_n-f|\leqslant \varepsilon\}$.

Conversely, if $d(f_n,f)\to 0$, then $\frac{\varepsilon}{1+\varepsilon}\mu\{|f_n-f|>\varepsilon\}\to 0$ (integrating over the set $\{|f_n-f|>\varepsilon\}$).