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
 A: 
Let $(X,S,\mu)$ is a measure space and $\mu(X)<\infty$. Define $d(f,g)=\int\frac{|f-g|}{1+|f-g|}d\mu$. Then $d$ is a metric on the space of measurable functions.

Let us prove a more general result.
Let $d: \Bbb R \times \Bbb R \rightarrow [0, +\infty) $ be any metric in $\Bbb R$.
Let us define $d_0: \Bbb R \times \Bbb R \rightarrow [0, 1) $ by
$$d_0(x,y) = \frac{d(x,y)}{1+d(x,y)}$$
Then $d_0$ is also a metric in  $\Bbb R$.
In fact

*

*$d_0(x,y) = 0$ iff $d(x,y) = 0$ iff $x=y$.

*$d_0(y,x) = \frac{d(y,x)}{1+d(y,x)}=  \frac{d(x,y)}{1+d(x,y)} = d_0(x,y)$

*For triangular inequality,
let $f: [0, +\infty) \rightarrow \Bbb [0,1)$  defined as $f(t) = \frac{t}{1+t}$. Note that $f'(t) = \frac{1}{(1+t)^2}>0$ so $f$ is strictly increasing. Note that $d_0= f \circ d$. So we have
\begin{align*} d_0(x,z)&= f(d(x,z))\leq f(d(x,y) +d(y,z)) = \frac{d(x,y) +d(y,z)}{1+d(x,y) +d(y,z)} = \\
 & = \frac{d(x,y) }{1+d(x,y) +d(y,z)}+ \frac{d(y,z)}{1+d(x,y) +d(y,z)} \leq \\
& \leq \frac{d(x,y) }{1+d(x,y)}+ \frac{d(y,z)}{1+d(y,z)} = d_0(x,y) + d_0(y,z)
\end{align*}
So, $d_0$ is a metric in $\Bbb R$ (actually  $d_0$ is equivalent to $d$, but we don't need to prove this for this question).

Now, let $\mathcal{M}(X,S)$ be the set of measurable functions defined in $(X,S)$.
Let $d_\mu: \mathcal{M}(X,S) \times \mathcal{M}(X,S) \rightarrow [0, \infty)$ be defined by
$$ d_\mu (f,g) =\int d_0(f(x),g(x)) d\mu(x) = \int \frac{d(f(x),g(x))}{1+d(f(x),g(x))}d\mu(x)$$
Note that, since, for all $x$, $d_0(f(x),g(x))\in [0,1)$ and $\mu(X) <+\infty$, we have that, for all $f, g \in \mathcal{M}(X,S)$, $d_\mu (f,g)  \leq \mu(X) <+\infty$. So, $d_\mu (f,g) \in [0, +\infty)$.
In order that $d_\mu$ be a metric, we must make the convention that, if $f,g \in \mathcal{M}(X,S)$ and $f=g$ a.e., we are going to consider that $f=g$ (technically we are taking the quotient of $\mathcal{M}(X,S)$  by the equivalence relation $=$ a.e.).
Let us prove that $ d_\mu$ is a metric.

*

*$d_\mu (f,g)  = 0$ iff $d_0(f(x),g(x))=0$ a.e  iff $f=g$ a.e.

*$ d_\mu (g,f) =\int d_0(g(x),f(x)) d\mu(x) = \int d_0(f(x),g(x)) d\mu(x)= d_\mu (f,g) $

*

\begin{align*} d_\mu (f,h) &= \int d_0(f(x),h(x)) d\mu(x) \leq \\ 
& \leq \int d_0(f(x),g(x)) d\mu(x) + \int d_0(g(x),h(x)) d\mu(x) =\\ 
& =  d_\mu (f,g)+ d_\mu (g,h)
\end{align*}
So, $d_\mu$ is a metric in $\mathcal{M}(X,S)$.
The result in the question is just the special case where $d(x,y)=|x-y|$.
Remark:
Note that in addition to the triangular inequality for $d_\mu$ (which depends on the triangular inequality for $d_0$), there are two other subtle points:

*

*We use the fact that $\mu(X)<\infty$ to ensure that $d_\mu$ is finite. Having $d_\mu$ finite is a condition for $d_\mu$, as defined in the question, to be a metric.


*If $f,g \in \mathcal{M}(X,S)$ and $f=g$ a.e., we must identify $f$ and $g$ (technically we are taking the quotient of $\mathcal{M}(X,S)$  by the equivalence relation $=$ a.e.). This is needed in order that $d_\mu$, as defined in the question, is a metric (otherwise $d_\mu$ would be only a pseudo-metric).
A: 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 $d$.

