Convergence in measure and metric notion

Any help with this problem is appreciated.

Given a measurable set of finite measure $D$, define $L_0(D)$ to be the vector space of real valued measurable functions on $D$. Define $d(f,g) := \int_D \min\{|f-g|,1\}$. $d(.,.)$ is a metric can be proven. I wanted to know how to prove the following statements, $$(f_n) \text{ is Cauchy in measure } \Leftrightarrow (f_n) \text{ is Cauchy in } L_0(D,d) \text { and }$$
$$ d(f_n,f) \rightarrow 0 \Leftrightarrow f_n \rightarrow f \text{ in measure}$$

I was able to show one side of $ d(f_n,f) \rightarrow 0 \Leftrightarrow f_n \rightarrow f$ in measure. The proof for that is as follows: For $\epsilon \in (0,1)$

$$(\Leftarrow) \, d(f_n,f) \leq m(D \cap \{\vert f_n - f \vert \geq \epsilon \}) + \epsilon m(D) \rightarrow \epsilon (1+m(D)) \text{ as n goes to } \infty $$ (Is this enough for $(\Leftarrow)$? and how to proceed with the rest?)

Answer

It's enough for $\Leftarrow$ since we get that for each $\varepsilon>0$, $\limsup_{n\to +\infty}d(f_n,f)\leqslant \varepsilon(1+m(D))$.

For the other direction, assume that $d(f_n,f)\to 0$ and fix $1>\varepsilon>0$. Then
$$\int_D\min\{1,|f_n(x)-f(x)|\}dm\geqslant \int_{\{|f_n(x)-f(x)|>\varepsilon\}}\min\{1,|f_n(x)-f(x)|\}dm\\\geqslant \varepsilon m\{|f_n(x)-f(x)|>\varepsilon\}$$