Build a function $f:[0,1] \rightarrow R$ that is unbounded and Cauchy-Integrable. 
Def.: A partition of an interval $[a,b]$ is a set of points  $ P=\{x_{0}, x_{1}, ..., x_{n}\}$, such that, $x_{0}=a<x_{1}<...<x_{n}=b.$
Def.: $|P|=max\{x_{i}-x_{i-1} : 1\leq i \leq n\}$.
Notation: $$ S(f,P)= \sum_{i=1}^{n}f(x_{i-1})(x_{i}-x_{i-1})$$
Def.: A function  $ f:[a,b] \rightarrow {R}$  is C-Integrable if  $ \exists $   l  $ \in {R} $  such that $ \forall \varepsilon >0$,  $ \exists \delta >0$ with $ \forall P, |P| < \delta \Rightarrow |S(f,P)-l| < \varepsilon.$

Sugestion of the teacher: 
$f(x)= \left
  \{
  \begin{array}{ll}
   n & \textrm{if }x=1-\frac{1}{n^3} \textrm{, for $n$ integer and } n \geq 2 \\
   0 & \textrm{otherwise }
   \end{array}
  \right. $
 A: (i)
$f$ is unbounded:
Supose that $f(x)$ is bounded. We must have $\exists M \in R$ such that $|f(x)|<M$
Take $x=1-\frac{1}{M^3} \in [0,1]$.  So, $f(x)=M$. Therefore $f$ can not be bounded.
(ii)
$f$ is C-Integrable:
As the series $\sum_{n=1}^\infty \frac{1}{n^2}$ is convergent, for all $\varepsilon>0$,  exists $n_0 \geq 1$ such that 
        $$
  \sum_{n>n_0}^{} \frac{1}{n^2}<\frac{\varepsilon}{2}
  $$
        Choose $\delta>0$ such that $\delta(1+2+...+n_0)<\frac{\varepsilon}{2}$ and be $P=\{t_0,...,t_k\}$ one partition of $[0,1]$ with $|P|<\delta$. Define $\Lambda=\{1,...,k-1\}$ and:
        $$
  \Lambda_1=\{ i \in \Lambda: t_i=1-\frac{1}{n^3}\textrm{  for } n=1,...,n_0\}
  $$
        $$
  \Lambda_2=\{ i \in \Lambda: t_i=1-\frac{1}{n^3}\textrm{ for } n>n_0\}
  $$
        So:
        $$
  S(f,P)= \sum_{i \in \Lambda_1}f(t_i)(t_{i+1}-t_i)+\sum_{i \in \Lambda_2}f(t_i)(t_{i+1}-t_i)
  $$
        Note that:
        $$
  \leq  \sum_{i \in \Lambda_1}f(t_i)(t_{i+1}-t_i) \leq \delta \sum_{i \in \Lambda_1}f(t_i) \leq \delta(1+2+...+n_0)<\frac{\varepsilon}{2}
  $$
And addition, if $i \in \Lambda_{2}$ then $ t_{1+i} - t_i < 1 - t_i = \frac{1}{n^3}$, for some integer $n>n_0$, therefore:
$$
0 \leq \sum_{i \in \Lambda_2}f(t_i)(t_{i+1}-t_i) \leq \sum_{n>n_0} \frac{1}{n^2} < \frac{\varepsilon}{2} $$
So $S(f,P)<\varepsilon$ and the function is C-Integrable.
