Showing if $f$ unbounded, then $f$ is not Riemann Integrable I want to show that if $f:[a,b] \to \mathbb{R}$ is unbounded, then $f$ is not Riemann Integrable. 
I suppose that $f$ is unbounded above:
$$\forall_{n \in \mathbb{N}} \exists_{t_n} \in [a,b], \; f(t_n) > n$$
and I note that if $f$ is integrable, then there exists two sequences $P, Q < \delta$ s.t. $|\sigma(f,P,\xi) - \sigma(f,Q, \zeta)| < \epsilon$ 
So if I want to show that if $f$ is not integrable, then I must negate the above statement:
$$\exists_{\epsilon > 0} \forall_{\delta > 0} ||P|| < \delta ||Q|| < \delta, \; |\sigma(f,P,\xi) - \sigma(f,Q,\zeta)| \geq \epsilon$$ where $P$ and $Q$ are sequences in $[a,b]$. 
I let $P = Q = \left \{ x_k\right \}_{k=0}$ and $\epsilon = 1$. However, it is here where I am stuck. How do I choose my sequence $\{x_k\}$?

EDIT: I think I can just pick a sequence with an even interval. i.e, like $x_k = a + k\left(\frac{b-a}{m}\right )$ where $m$ is the number of subintervals. Then we have that $|\Delta x_k| = \frac{b-a}{m}$

We can assume that $f$ is unbounded above without loss of generality.
Then there are $x_n$ such that $f(x_n) \ge n$, and since $[a,b]$ is compact,
there is some accumulation point $c$ such that $\sup_{x \in (c-\epsilon, c+\epsilon)\cap[a,b]} f(x) = \infty$ for all $\epsilon>0$.
Let $P$ be a partition of $[a,b]$. Suppose $c \in I^\circ$, where $I$ is an interval of the partition. Then we see that
$\sup_{x \in I} f(x) = \infty$.
We see that if $c$ is an endpoint, we obtain the same result.
If $c$ lies on the boundeary between two intervals, at least one of the intervals must satisfy $\sup_{x \in I} f(x) = \infty$.
In all cases, we see that $U(f,P) = \infty$, hence $f$ is not integrable.
