Let $m(E) < \infty$. For the measurable functions $g$ and $h$ on $E$, define
$\rho(g,h) = \int_E \frac{|g-h|}{1+|g-h|}$. , where $\int$ denotes the lebesgue integral.
Show that $\{f_n\} \rightarrow f $ in measure if and only if $lim_{n\rightarrow\infty} \; \rho (f_n,f) = 0$.
Intuition suggests that one should probe this by considering the case where $f_n$ does not converge in measure to $f$ , but I'm not sure how to do this.
If $f_n$ doesn't converge in measure to $f$, then there is an $\epsilon>0$ and $\delta>0$ and subsequence $f_{n_k}$ such that
$$m \left( \{ x : |f_{n_k}(x)-f(x)| \geq \epsilon \} \right) \geq \delta.$$
Set $E_k=\{ x : |f_{n_k}(x)-f(x)| \geq \epsilon \}$ and set $g(x)= \frac{x}{1+x}$. Use the fact that $g(x)$ is increasing for $x \geq 0$ and monotonicity of the integral to show $\rho( f_{n_k},f)$ doesn't converge to zero.