Vector space of all Lebesgue measurable functions Let $L_0$ be the vector space of all Lebesgue measurable functions on $[0,1]$ with metric  $d(f,g)$ = $\int_{0}^{1} |f(t)-g(t)|/( 1+ |f(t)-g(t)| ) dt $ .
Show that $d(f_n,f) \to 0$ iff $f_n \to f$ in measure and that $(L_0,d)$ is a topological vector space.
I know that ${f_n} $ converges in measure to a measurable function $f$  if  $ lim  [  \mu \{ |f_n - f| \ge \epsilon \}]=0$ for all choices of $\epsilon > 0$.But I am unable to solve this problem using the definition. Need some help.
 A: Let's first assume convergence in measure. For $\epsilon > 0$, define the set $A_{n} := \{t \in [0,1] \; :\;|f_{n}(t) - f(t)| < \epsilon\}$. We obtain
$$d(f_{n}, f) = \int_{A_{n}} \frac{|f_{n}(t) - f(t)|}{1 + |f_{n}(t) - f(t)|}\; dt + \int_{A_{n}^{c}} \frac{|f_{n}(t) - f(t)|}{1 + |f_{n}(t) - f(t)|}\; dt$$
Now the first integral is smaller than $\epsilon$ by definition of $A_{n}$. For the second integral, observe that the integrand is always smaller than $1$, hence we can make the estimate
$$d(f_{n},f) \leq \epsilon + \mu(A_{n}^{c}) = \epsilon + \mu(\{t: |f_{n}(t) - f(t)| \geq \epsilon \})$$ 
EDIT: For the converse, let's try showing the contraposition: Assume $f_{n}$ does not converge in measure against $f$, then there must be an $\epsilon > 0$ and a $\delta > 0$ with $\mu(\{t : |f_{n_k}(t) - f(t)| \geq \epsilon\}) > \delta$ for some subsequence $n_{k}$. Using that the function $x \mapsto \frac{x}{1+x}$ is monotonically increasing on $\mathbb{R}_{>0}$, we obtain the estimate (Let $A_{n_{k}}$ be defined as above)
$$d(f_{n_{k}},f) \geq \int_{A_{n_{k}}^{c}} \frac{|f_{n_{k}}(t) - f(t)|}{1 + |f_{n_{k}}(t) - f(t)|}\; dt \geq \frac{\epsilon}{1 + \epsilon}\cdot \mu(A_{n_k}^{c}) > \frac{\epsilon}{1+\epsilon}\delta$$
which proves that $d(f_{n},f)$ can not converge to $0$.
The continuity of addition and multiplication follows easily from the properties of convergence in measure.
