$d(f, g)=\int\frac{|f-g|}{1+|f-g|}d\mu$ and Cauchy sequences

Question

Let $(X,S,\mu)$ be a measure space with $\mu(X)<\infty$ and define for each $f\in\mathbb{M}(X,S)$ ($f$ measurable):

$$r(f):=\int\frac{|f|}{1+|f|}d\mu$$

I showed that:

$r(f)<\infty$

$d(f,g):=r(f-g)$ is a metric

$d(f_n,f)\xrightarrow[]{n\to\infty}0\Leftrightarrow f_n\xrightarrow[\mu]{n\to\infty}f$ (Pointwise convergence iff convergence in measure)

Now I would like to demonstrate completeness. I want to show that if $(f_n)$ is a Cauchy sequence with respect to $d$ in $\mathbb{M}(X, S)$, then there exists $f\in\mathbb{M}(X,S)$ such that $d(f_n,f)\xrightarrow[]{n\to\infty}0$.

I think the result is a simple consequence of the F. Riesz - H. Weyl theorem but I'm not entirely sure.

Answer 1

$d(f_n,f_m) \geq \int_{|f_n-f_m| >\epsilon} \frac {|f_n-f_m|}{1+|f_n-f_m|} d\mu \geq \frac {\epsilon} {1+\epsilon} \mu (|f_n-f_m| >\epsilon)$. Hence $(f_n)$ Is Cauchy in measure. This implies that it converges in measure. For a proof of this fact see Cauchy in measure implies convergent in measure.

Let $f_n\rightarrow f$ in measure. That this implies $d(f_n,f) \to 0$ is immediate from one form of DCT where a.e. convergence is replaced by convergence in measure. This form of DCT is proved by just going to subsequences and using the standard form of DCT.