As an example that not every Cauchy sequence in $(M,d)$ is converging in $M$ the following examples are given:
Consider $(\mathbb{Q},d_{\text{eucl}})$ and a sequence $q_n \in \mathbb{Q}\to \pi$. Which is a Cauchy sequence (as $d_\text{eucl}(q_1,q_2)\to 0)$ but the sequence does not converge in $\mathbb{Q}$.
A second example is given. Consider: $$\left(\mathcal{C}([0,1],\mathbb{R}), d_1\right) \qquad \text{and} \qquad d_1(f,g) = \int_0^1 |f-g|$$ Where $f \in \mathcal{C}([0,1],\mathbb{R})$ mean $f: [0,1]\to \mathbb{R}$ continous.
I need to find a counterexample in this last metric space. I know I need to search for a sequence of functions $(f_n)_n$ where $d_1(f_a,f_b)\to 0$ but the limit shouldn't live in that space of functions. How can I find this?
Note
Below the problem is written: 'in $\left(\mathcal{L}^1([0,1],\mathbb{R}), d_1\right)$ every Cauchy sequence will be converging.'
Where $\mathcal{L}^1([0,1],\mathbb{R})$ is the space of Lesbeque integrable functions. I guess I should construct a sequence of functions wich converges to a function which is Lesbeque integrale, but not continous. Could someone help?
Consider the functions $f_n(x)$ where they take on value $0$ for $x< \frac{1}{2} - \frac{1}{n}$, $1$ for $x > \frac{1}{2} + \frac{1}{n}$ and grow linearly from $0$ to $1$ on $\frac{1}{2} - \frac{1}{n} < x < \frac{1}{2} + \frac{1}{n}$.
Then, $d(f_n, f_m) \leq 4 \frac{1}{n}$ for $n <m$ (since each $f_n$ is bounded between $0$ and $1$ and the two functions only differ on $\frac{1}{2} - \frac{1}{n} < x < \frac{1}{2} + \frac{1}{n}$). So, the sequence is Cauchy.
However, the limit of $f_n$ with respect to this would be $0$ for $x<\frac{1}{2}$ and $1$ for $x > \frac{1}{2}$, which is not continuous.