TITLE: Showing $(C[0,1], d_1)$ is not a complete metric space
QUESTION [12 upvotes]: I am completely stuck on this problem: $C[0,1] = \{f: f\text{ is continuous function on } [0,1] \}$ with metric $d_1$ defined as follows:
$d_1(f,g) = \int_{0}^{1} |f(x) - g(x)|dx $.
Let the sequence $\{f_n\}_{n =1}^{\infty}\subseteq C[0,1]$ be defined as follows:
$
   f_n(x) = \left\{ 
  \begin{array}{l l}
     -1 & \quad \text{ $x\in [0, 1/2 - 1/n]$}\\
     n(x - 1/2) & \quad \text{$x\in [1/2 - 1/n, 1/2 +1/n]$}\\
  1 & \quad \text{ $x\in [1/2 +1/n, 1]$}\\
   \end{array} \right.
 $
Then $f_{n}$ is cauchy in $(C[0,1], d_1)$ but not convergent in $d_1$.
I have proved that $f_{n}$ is not convergent in $(C[0,1])$ since it is converging to discontinuous function given as follows:
$
   f_n(x) = \left\{ 
  \begin{array}{l l}
     -1 & \quad \text{ $x\in [0, 1/2 )$}\\
     0 & \quad \text{$x = 1/2$}\\
  1 & \quad \text{ $x\in (1/2 , 1]$}\\
   \end{array} \right.
 $
I am finding it difficult to prove that $f_{n}$ is Cauchy in $(C[0,1], d_1)$. 
I need help to solve this problem.
Edit: I am sorry i have to show $f_n$ is cauchy
Thanks for helping me.

REPLY [12 votes]: Suppose $m,n > N$. Then $f_m(x) = f_n(x) = -1 $ when $x \in [0, \frac{1}{2}-\frac{1}{N}]$. Similarly, $f_m(x) = f_n(x) = +1 $ when $x \in [\frac{1}{2}+\frac{1}{N},1]$. And $|f_m(x)-f_n(x)| <
 1$ when $x \in (\frac{1}{2}-\frac{1}{N}, \frac{1}{2}+\frac{1}{N})$.
Hence $d_1(f_m,f_n) = \int_{0}^{1} |f_m(x) - f_n(x)|dx = \int_{\frac{1}{2}-\frac{1}{N}}^{\frac{1}{2}+\frac{1}{N}} |f_m(x) - f_n(x)|dx < \frac{2}{N}$.