Show that $f$ is bounded if there is sum of Riemann. *

Let be $f:A\longrightarrow\mathbb{R}$ a any function, defined in block
  $A$. If there is a limit $\lim_{|P|\rightarrow 0} \sum(f;P*)$ then f
  is limited.

*
My proof: 
Suppose $f$ unlimited, by contradiction. Then take a sequence of blocks $B_1, ... , B_k \in P$. Hence, since we suppose $f$ unlimited there exists at least one $B_{\xi}$ such that $|f(x_{\xi})|\ge M$, where $M$ any constant.
So we have to $\sum(f;P*)=\sum_{i=1, i\ne \xi}^k(f(x_i)vol.B_i) + f(x_{\xi})vol.B_{\xi}$. Then, as the second portion diverges and the first converges, then the sum diverges. Thus we get the result.
Sorry for my bad English. I gave a demonstration to this problem but I discovered a mistake, so I decided to ask for help for you.
Thanks in advance.
 A: You need to show by contradiction that if $f$ is unbounded in $A$, then it is impossible for the limit of Riemann sums, $\lim_{|P| \to 0} S(f,P)$, to attain a finite limit.  
Assume there exists $\epsilon_0 >0$ such that for any number $J \in \mathbb{R}$ and any $\delta >0$ there exists a partition $P$ with $|P| < \delta$ and 
$$|S(P,f) - J| =|\sum_{i \neq \xi} f(x_i) vol(B_i) + f(x_\xi)vol(B_\xi) - J|  > \epsilon _0.$$
Just choose $\epsilon_0 = 1$ and given $\delta > 0$ choose any partition with $|P | < \delta$.  As you stated, there exists at least one index $\xi$ such that $f$ is unbounded on $B_\epsilon$.  
Now select $x_\xi \in B_\xi$ such that
$$|f(x_\xi)| > \frac{1 + | J - \sum_{i \neq \xi} f(x_i) vol(B_i)|}{vol(B_\xi)}.$$
Using the reverse triangle inequality we have
$$|\sum_{i \neq \xi} f(x_i) vol(B_i) + f(x_\xi)vol(B_\xi) - J| \geqslant |f(x_\xi)|vol(B_\xi) - |J - \sum_{i \neq \xi} f(x_i) vol(B_i)| > 1.$$
This contradicts the existence of the limit so we must have $f$ bounded.
