Let $L^0([0,1])$ be the vector space of Lebesgue-measurable functions on $[0,1]$. Let $d$ be the metric on $L^0([0,1])$ given by $$d(f,g) = \int_0^1 \frac{|f(x)-g(x)|}{1 + |f(x)-g(x)|}\, dx.$$Prove that $f_n \xrightarrow{d} f$ (convergence in the metric) if and only if $f_n \xrightarrow{m} f$ (convergence in measure.)
My work. Suppose $f_n \xrightarrow{m} f$ and fix $\varepsilon > 0$. Now,$$\begin{align*}d(f_n,f) = \int_0^1 \frac{|f_n(x)-f(x)|}{1 + |f_n(x)-f(x)|}\, dx = \int_{\{|f_n-f| > \varepsilon\}} \frac{|f_n(x)-f(x)|}{1 + |f_n(x)-f(x)|}\, dx + \int_{\{|f_n-f| \leq \varepsilon\}} \frac{|f_n(x)-f(x)|}{1 + |f_n(x)-f(x)|}\, dx.\end{align*}$$We can bound the second integral as$$\int_{\{|f_n-f| \leq \varepsilon\}} \frac{|f_n(x)-f(x)|}{1 + |f_n(x)-f(x)|}\, dx \le \varepsilon \int_{\{|f_n-f| \leq \varepsilon\}} \frac{1}{1 + |f_n(x)-f(x)|}\, dx \le \varepsilon\int_0^1\, dx = \varepsilon.$$For the first integral,$$ \int_{\{|f_n-f| > \varepsilon\}} \frac{|f_n(x)-f(x)|}{1 + |f_n(x)-f(x)|}\, dx \le \int_{\{|f_n-f| > \varepsilon\}}\, dx = m(\{|f_n-f| > \varepsilon\}).$$In total,$$d(f_n,f) \le \varepsilon + m(\{|f_n-f| > \varepsilon\}).$$So, $$0\le \limsup_{n\to\infty} \, d(f_n,f) \le \varepsilon + \limsup_{n\to\infty} m(\{|f_n-f| > \varepsilon\}) = \varepsilon.$$Since $\varepsilon > 0$ is arbitrary, $\displaystyle\limsup_{n\to\infty} \, d(f_n,f) = 0$, i.e., $\displaystyle\lim_{n\to\infty} \,d(f_n,f) = 0$.
For the other direction, suppose $f_n \xrightarrow{d} f$, i.e., $$\lim_{n\to\infty}\int_0^1 \frac{|f_n(x)-f(x)|}{1 + |f_n(x)-f(x)|}\, dx = 0.$$I'm not sure where to go next, to show $f_n \xrightarrow{m} f$. Thanks for any help!