measure convergence Let $(X,\mathcal{F},\mu)$ be a finite measurable space. Define $$d(f,g) = \int_X \frac{|f(x)-g(x)|}{1+|f(x)-g(x)|}\mu(dx) $$
Proof that $d(f_n,f)\to0 \Leftrightarrow\ f_n$ converge in measure to $f$
Attempt: 
I did proof that d is distance $(d>0,\space d(f,g)=d(g,f),\space d(f,g)<d(f,h)+d(h,g))$ not that I'm sure it helped. In any case, as far as I know, converging in measure means that for every $\epsilon$,$\lim_{n\to\infty}\mu(\{x\in X :|f(x)-f_n(x)|>\epsilon\})=0$
for the $\Rightarrow$ part, I am thinking of applying lebesgue's dominated convergence since the integrand is bounded by 1 and since the space is finite, 1 will be integrable. But then, just because the value of $\int_X\lim_{n\to\infty}(1-\frac{1}{1+|f(x)-g(x)|})=0$ doesnt mean that the inside must be 0 a.e 
for the $\Leftarrow$, I honestly have no idea how to manipulate that into the distance function
Any help is appreciated. Thank you
 A: Define $D(a,b) = |a - b|/(1 + |a - b|)$, for all $a,b\in \Bbb R$. To prove the forward direction, assume $d(f_n,f) \to 0$. Let $\epsilon > 0$. Since the function $x\mapsto \frac{x}{1 + x}$ is strictly increasing on $[0, \infty)$, 
$$\mu\{x\in X :|f_n(x) - f(x)| > \epsilon\} \le \mu\left\{x\in X :D(f_n(x),f(x)) > \frac{\epsilon}{1 + \epsilon}\right\}.$$
Applying Chebyshev's inequality to the last expression, and using the assumption $d(f_n,f)\to 0$, we find that  
$$\mu\left\{x\in X : |f_n(x) - f(x)| < \epsilon\right\} \le \frac{1 + \epsilon}{\epsilon} \int D(f_n, f)\, d\mu = \frac{1+\epsilon}{\epsilon}\rho(f_n,f) \to 0.$$
Hence $\mu\{x\in X :|f_n - f| > \epsilon\} \to 0$, and consequently $f_n$ converges in measure to $f$.
The assumption $\mu(X) < \infty$ is mecessary to prove the reverse direction. Assume $f_n$ converges in measure to $f$. Let $\epsilon > 0$. Define $A_n(\epsilon) := \{x\in X :|f_n - f| > \epsilon\}$, for all $n\in \Bbb N$. Then 
$$d(f_n,f) = \int_{A_n(\epsilon)} D(f_n,f)\, d\mu + \int_{X\setminus A_n(\epsilon)} D(f_n,f)\, d\mu \le \mu(A_n(\epsilon)) + \int_{X\setminus A_n(\epsilon)} D(f_n,f)\, d\mu.$$
Since $D(f_n,f) \le \frac{\epsilon}{1 + \epsilon}$ on $X\setminus A_n(\epsilon)$, 
$$\int_{X\setminus A_n(\epsilon)} D(f_n,f)\, d\mu \le \frac{\epsilon}{1 + \epsilon}\mu(X\setminus A_n(\epsilon)) \le \frac{\epsilon}{1 + \epsilon}\mu(X).$$
Therefore
$$\limsup_{n\to \infty} d(f_n,f) \le \limsup_{n\to \infty}  \left(\mu(A_n(\epsilon)) + \frac{\epsilon}{1 + \epsilon}\mu(X)\right) \le \frac{\epsilon}{1 + \epsilon}\mu(X),$$
since $\mu(A_n(\epsilon)) \to 0$ by assumption. As $\mu(X) < \infty$ and $\epsilon$ was arbitrary, $d(f_n,f) \to 0$.
