Let $([0,1], \mathscr{B}([0,1]), \lambda)$ measure space. If $f,g \in M([0,1],\mathscr{B}([0,1]))$ we define the pseudometric $$d(f,g)= \int_{0}^{1}\min\{|f(x)-g(x)|,1\}d\lambda$$
Prove if $(f_{n})_{n \in \mathbb{N}}$ a sequence functions measurable is $d-$Cauchy then is Cauchy measure. *Note: $d-$ Cauchy means, $\forall \epsilon>0, \exists N \in \mathbb{N}$ such that if $n,m \geq N$ then $d(f_n,f_m)<\epsilon$
We know that $(f_{n})_{n \in \mathbb{N}}$ is $d-$Cauchy, $\exists N \in \mathbb{N}$ then $\mu(\{x \in [0,1]: d(f_n(x),f_m(x)) \geq \epsilon \})=0$ but just I have $$\{x \in [0,1]: d(f_n(x),f_m(x)) \geq \epsilon \} \subset \{x \in [0,1]: |f_n(x)-f_m(x)| \geq \epsilon \}$$ So, do you know other relation with that sets or other form to get the solution?
Let $0<\varepsilon <1$. Then $$\begin{aligned}\lambda(|f_n-f_m|>\varepsilon)&=\int_{\{\varepsilon<|f_n-f_m|< 1\}}d\lambda + \int_{\{|f_n-f_m|\geq 1\}}d\lambda=\\ &= \varepsilon^{-1}\int_{\{\varepsilon<|f_n-f_m|< 1\}}\varepsilon d\lambda + \int_{\{|f_n-f_m|\geq 1\}}\min(|f_n-f_m|,1)d\lambda\leq \\ &\leq \varepsilon^{-1}\int_{\{\varepsilon<|f_n-f_m|< 1\}}\min(|f_n-f_m|,1)d\lambda + \int_{\{|f_n-f_m|\geq 1\}}\min(|f_n-f_m|,1)d\lambda\leq\\ &\leq (1+\varepsilon^{-1})\int_{[0,1]}\min(|f_n-f_m|,1)d\lambda \end{aligned}$$ Let $\varepsilon \geq 1$. Then $$\begin{aligned}\lambda(|f_n-f_m|>\varepsilon)&=\int_{\{|f_n-f_m|>\varepsilon\}}d\lambda=\\ &=\int_{\{|f_n-f_m|>\varepsilon\}}\min(|f_n-f_m|,1)d\lambda\leq\\ &\leq \int_{[0,1]}\min(|f_n-f_m|,1)d\lambda \end{aligned}$$