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 $\{|f_n-f|>\varepsilon\}$).
(a) The integrand $\frac{|f-g|}{1+|f-g|}$ is nonnegative for all $x\in X$ because $|f-g|$ is nonnegative and adding $1$ ensures the denominator is positive, making the fraction nonnegative. Therefore, the integral of a nonnegative function over a measure space is nonnegative. Obviously, $d(f,g)=0$ if $f=g$, however, if $f=g,\text{a.e. }x\in X$, then $|f-g|=0\text{ a.e. }x\in X$ which means $d(f,g)=0$. Therefore, $d$ is a semimetric, but not a metric. Indeed, $d(f,g)=0$ if and only if $f=g,\text{ a.e. }x\in X$. $$ d(f,g)=\int\frac{|f-g|}{1+|f-g|}\,\mathrm{d}\mu=\int\frac{|g-f|}{1+|g-f|}\,\mathrm{d}\mu=d(g,f) $$ To show the triangle inequality, consider the properties of the function $h(t)=\frac{t}{1+t}$. It is an increasing function for $t\ge 0$ and satisfies $h(t_{1}+t_{2})\le h(t_{1})+h(t_{2})$ for $t_{1},t_{2}\ge 0$ by direct calculation. Thus, for any $x$, we have $$ \frac{|f-h+h-g|}{|1+|f-h+h-g|}\le\frac{|f-h|+|h-g|}{1+|f-h|+|h-g|}\le\frac{|f-h|}{1+|f-h|}+\frac{|h-g|}{1+|h-g|} $$ Integrating both sides over $X$ with respect to $\mu$ gives us $d(f,g)\le d(f,h)+d(h,g)$, which confirms the triangle inequaliaty.
(b)If $f_{n}\to f$ in measure, fix $\varepsilon$ and integrate over $\{x\in X:|f_{n}(x)-f(x)|>\varepsilon\}$ and $\{x\in X:|f_{n}(x)-f(x)|\le\varepsilon\}$, \begin{aligned} \int\frac{|f-f_{n}|}{1+|f-f_{n}|}\,\mathrm{d}\mu&\le\int 1\,\mathrm{d}\mu\\ &=\int_{\{x\in X:|f_{n}(x)-f(x)|>\varepsilon\}}1\,\mathrm{d}\mu+\int_{\{x\in X:|f_{n}(x)-f(x)|\le\varepsilon\}}1\,\mathrm{d}\mu\\ &\le\mu\Big(\{x\in X:|f_{n}(x)-f(x)|>\varepsilon\}\Big)+\varepsilon\mu(X). \end{aligned} Thus $d(f_{n},f)\to 0$ as $n\to\infty$.
Conversely, If $d(f_{n},f)\to 0$ as $n\to\infty$, then \begin{aligned} \int_{X}f\,\mathrm{d}\mu&\ge\int_{\{x\in X:|f_{n}(x)-f(x)|>\varepsilon\}}f\,\mathrm{d}\mu\\ &\ge\frac{\varepsilon}{1+\varepsilon}\mu\Big(\{x\in X:|f_{n}(x)-f(x)|>\varepsilon\}\Big) \end{aligned} It follows that $f_{n}\to f$ in measure.
