Convergence in measure metrizable?

Question

I am trying to show for a $\sigma$-finite measure space $f_n\rightarrow f$ $\mu$-stochastically iff $\lim_{n\rightarrow\infty} d(f_n,f)=0$ where $$d(f,g):=\sum^\infty_{k=1}\frac{2^{-k}}{\mu(\Omega_k)}\int I_{\Omega_k}\frac{|f_n-f|}{1+|f_n-f|}d\mu$$ but I'm having trouble proving either direction. For the forward direction I say from the definition (w.l.o.g. replacing $f_n$ by $f_n-f$): $$\forall A\in\mathcal{A},\epsilon>0:\mu(A\cap\{f_n>\epsilon\})\rightarrow 0 (n\rightarrow\infty),$$ it follows that $$\forall A\in\mathcal{A},\epsilon>0:\mu(A\cap\{\frac{f_n}{1+f_n}>\frac{\epsilon}{1+\epsilon}\})\rightarrow 0 (n\rightarrow\infty).$$ Then by choosing a constant some $C=1/(\epsilon+1)>0$ argue as follows $$d(f_n,0):= \sum^\infty_{k=1}\frac{2^{-k}}{\mu(\Omega_k)}\int I_{\Omega_k}\frac{|f_n|}{1+|f_n|}d\mu$$ $$\leq\sum c_k \int I_{\Omega\cap\{|f_n|/(1+|f_n|)>C\}}\leq \sum c_k \mu(A\cap\{\frac{|f_n|}{1+|f_n|}>C\})\rightarrow 0(n\rightarrow\infty).$$ Is this the right approach? Is there such a constant? For the other direction I argue for a non-empty measurable set $A$ with finite measure and $\epsilon,\delta>0$ we can choose $A$ to be one of the partitioning sets and it follows that for some $N\in\mathbb{N}$ and all $n\geq N$: $$\delta>d(f_n,0)\geq \int_{A}\frac{|f_n|}{2\mu(A)(1+|f_n|)}\geq \mu(A\cap\{\frac{|f_n|}{(1+|f_n|)}>\epsilon\})/2\mu(A).$$ And now the direction follows with $\delta\rightarrow 0$. Is this legal? Any help would be great! Sorry for the sloppy tex.

Answer 1

The metric $d(f,g)=\min\{\inf\limits_{\delta>0} \mu(\{|f-g|\geq \delta\})+\delta, 1\}$ does what you want, the one you defined only captures local convergence in measure.