Q) Let $\mu(X)<\infty$ and $V$ be the space of all measurable functions on $X$. Let $V_0 = \{f\in V| f=0, \mu \text{ -a.e. }\}$ and
$$d(f,g)= \int_X\frac{|f-g|}{1+|f-g|}d\mu$$
Show that $V/V_0$ is a complete metric space.
I've shown that $d$ is a metric but to show that $V/V_0$ is complete, I have to show that every Cauchy sequence converges to a limit in $V/V_0$ in metric $d$ but am not sure how to show that?
It suffices to show the completeness for $V$.
Since the metric on $V$ is invariant, the completeness for $V$ is equivalent to showing that \begin{align*} \sum_{n}\|f_{n}\|<\infty~~~~\rightarrow\sum_{n}|f_{n}|\in V. \end{align*} First we look at the inequality \begin{align*} \dfrac{|f|+|g|}{1+|f|+|g|}=\dfrac{|f|}{1+|f|+|g|}+\dfrac{|g|}{1+|f|+|g|}\leq\dfrac{|f|}{1+|f|}+\dfrac{|g|}{1+|g|}, \end{align*} then it is easy to see that \begin{align*} \dfrac{\displaystyle\sum_{n}|f_{n}|}{1+\displaystyle\sum_{n}|f_{n}|}\leq\sum_{n}\dfrac{|f_{n}|}{1+|f_{n}|}, \end{align*} and hence \begin{align*} \int_{X}\dfrac{\displaystyle\sum_{n}|f_{n}|}{1+\displaystyle\sum_{n}|f_{n}|}d\mu\leq\int_{X}\sum_{n}\dfrac{|f_{n}|}{1+|f_{n}|}d\mu=\sum_{n}\|f_{n}\|<\infty, \end{align*} we are done.