convergence of sequence of function in measure space

Question

Let $(X,\mathcal{M},\mu)$ be a finite measure space. Prove that $(f_n)_n$ converges to $f$ in measure if and only if $\mathrm{d}(f_n,f)\rightarrow 0$, where $\mathrm{d}$ is the metric defined as follows: given $f,g\,\colon X\to\mathbb{C}$, then $$ \mathrm{d}(f,g)=\int_X {|f-g|\over 1+|f-g|}\,d\mu. $$

Answer 1

Suppose that $f_n \to f$ in measure. Fix $\epsilon \gt 0$. Let $$E_{n,\frac{\epsilon}{2\mu(X)}}=\{x \in X: |f_n(x)-f(x)|\ge \frac{\epsilon}{2\mu(X)}\}$$ Then for every $\delta \gt 0$, there is a $n_0 \in \mathbb{N}$ such that $\forall n \ge n_0, \mu(E_{n,\frac{\epsilon}{2\mu(X)}}) \lt \delta$. Let $\delta=\frac{\epsilon}{2}$.Then for all $n \ge n_0$ $$\rho(f_n,f)=\int_{X}\frac{|f_n-f|}{1+|f_n-f|}d\mu=\int_{E_{n,\frac{\epsilon}{2\mu(X)}}}\frac{|f_n-f|}{1+|f_n-f|}d\mu+\int_{{E^c_{n,\frac{\epsilon}{2\mu(X)}}}}\frac{|f_n-f|}{1+|f_n-f|} d\mu \le \int_{E_{n,\frac{\epsilon}{2\mu(X)}}}1 d\mu+\int_{E^c_{n,\frac{\epsilon}{2\mu(X)}}}|f_n-f| d\mu\le \mu(E_{n,\frac{\epsilon}{2\mu(X)}})+\frac{\epsilon}{2\mu(X)}(\mu(X)) \lt \epsilon$$

Now suppose that $\rho(f_n,f) \to 0$. Then Let $$E_{n,\epsilon}=\{x \in X: |f_n(x)-f(x)|\ge \epsilon\}$$

Let $\delta \gt 0$. Then $$\rho(f_n,f)=\int_{X}\frac{|f_n-f|}{1+|f_n-f|}d\mu=\int_{E_{n,\epsilon}}\frac{|f_n-f|}{1+|f_n-f|}d\mu+\int_{{E^c_{n,\epsilon}}}\frac{|f_n-f|}{1+|f_n-f|} d\mu \ge \int_{E_{n,\epsilon}}\frac{|f_n-f|}{1+|f_n-f|}d\mu \ge \int_{E_{n,\epsilon}}\frac{\epsilon}{1+\epsilon}=\frac{\epsilon}{1+\epsilon}\mu(E_{n,\epsilon})$$ For the last inequality, observe that if $F(x)=\dfrac{x}{1+x}$, then $F$ is a strictly increasing function. Since for $x \in E_{n,\epsilon}, |f_n(x)-f(x)| \ge \epsilon, F(|f_n(x)-f(x)|) \ge F(\epsilon) $, which gives the desired inequality.

Now we have $$ \frac{\epsilon}{1+\epsilon} \mu(E_{n,\epsilon}) \le \rho(f_n,f)$$ There is a $n_0 \in \mathbb{N}$ such that for all $n \ge n_0$, we have $\rho(f_n,f) \lt \delta\left(\dfrac{\epsilon}{1+\epsilon}\right)$. Thus for all $n \ge n_0$, we have $\mu(E_{n, \epsilon}) \lt \delta$.