$f_n(x)$ convergence in measure implies $\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}$ convergence almost everywhere Let $E\subset\mathbb{R},m(E)<+\infty$, $\{f_n(x)\}$ are measurable functions defined on $E$. Then $\{f_n(x)\}$ converges to $f(x)$ in measure $\Leftrightarrow$ $$\lim_{n\rightarrow\infty}\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}=0, a.e. x\in E.$$
I think it's easy to see how to go from right to left since $\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}<\epsilon\Rightarrow|f_n(x)-f(x)|<2\epsilon$ and almost everywhere convergence implies convergence in measure.
What baffles me is the other direction.
 A: I guess the result you have to show is that the convergence in measure of $\{f_n\}$ to $f$ is equivalent to $$\lim_{n\to\infty}\int_E\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}dm (x).$$ 
(your result is not true, because $\displaystyle\lim_{n\to\infty}\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}$ is equivalent to $\lim\limits_{n\to\infty}|f_n(x)-f(x)|=0$ and there are sequences which converge in measure but not almost everywhere.)
To show the result, if $f_n\to f$ in measure, then fix $\varepsilon>0$. Since $t\mapsto \frac t{t+1}$ is increasing
$$\int_E\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}dm (x)\leq \frac{\varepsilon}{1+\varepsilon}m(E)+m(|f_n-f|\geq \varepsilon),$$
so $$\limsup_{n\to\infty}\int_E\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}dm (x)\leq \frac{\varepsilon}{1+\varepsilon}m(E),$$
and  $$\lim_{n\to\infty}\int_E\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}dm (x)=0.$$
Conversely, $$\frac{\varepsilon}{1+\varepsilon} \cdot m(|f_n-f|\geq\varepsilon)\leq \int_{\left\{|f_n-f|\geq\varepsilon\right\}}\frac{\varepsilon}{1+\varepsilon}dm(x)\leq \int_{\left\{|f_n-f|\geq\varepsilon\right\}}\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}dm(x),$$
and we are done.
