Let $(X,S,\mu)$ is a measure space and $\mu(X)<\infty$. Define $d(f,g)=\int\frac{|f-g|}{1+|f-g|}d\mu$ is a metric on the space of measurable functions Let $(X,S,\mu)$ is a measure space and $\mu(X)<\infty$. Define $d(f,g)=\int\frac{|f-g|}{1+|f-g|}d\mu$ is a metric on the space of measurable functions.
My work-
symmetry,
let's show $d(f,g)=d(g,f)$
$d(f,g)=\int\frac{|f-g|}{1+|f-g|}d\mu$ and $d(g,f)=\int\frac{|g-f|}{1+|g-f|}d\mu$
It's obvious that, $d(f,g)=d(g,f)$
My concern is triangular inequality,
What want to show is,
$d(f,h)\leq d(f,g)+d(g,h)$
so the right hand side becomes,
$\int\frac{|f-g|}{1+|f-g|}d\mu+\int\frac{|g-h|}{1+|g-h|}d\mu= \int\frac{|f-g|}{1+|f-g|}+\frac{|g-h|}{1+|g-h|}d\mu$
I don't think simplification of right side will give the result easily. Can someone give me a hints. Thank you in advance

The only thing missing is to check that
$$
\rho(x,y)=\frac{|x-y|}{1+|x-y|}
$$
is a metric on $\mathbb{R}$. Here is a simple proof:
Consider the function
$$ f(t)=\frac{t}{1+t}, \qquad t\geq0$$
notice that $f(t)=0$ iff $t=0$, $f$ is monotone non decreasing on $[0,\infty)$, and  that $\rho(x,y)=f(|x-y|)$.
That $\rho$ satisfies the triangle inequality is a consequence of
$$ f(t+s)\leq f(t)+f(s),\qquad t,s\geq0$$
which follows from
\begin{align}
f(t+s)&=\frac{s+t}{1+s+t}=\frac{t}{1+t}\frac{1+t}{1+t+s} +\frac{s}{1+s}\frac{1+s}{1+t+s}\\
&\leq \frac{t}{1+t}+\frac{s}{1+s}
\end{align}

Though the is not needed for the conclusions in your problem, it is worthwhile noticing that on any matrix space $(S,d)$, any continuous monotone nondecreasing function $\phi:[0,\infty)\rightarrow[0,\infty)$ such that

*

*$\phi(t)=0$ iff $t=0$,

*$\phi(t+s)\leq \phi(t)+\phi(s)$ (subadditve)

induces another  metric on $S$
$$\rho_\phi(x,y)=\phi(d(x,y))$$
that is equivalent to $d$.

