Proving a function is bounded Hi i was wondering if  anyone could help me with my revision This is a question of a past paper i'm stuck on.
Let $$f:[a,b]\to \Re$$ be Riemann intergrable. Part (a) is to prove that f is bounded and part (b) is to Give an example of a bounded function f that is not Riemann integrable
my attempt for part (a) is since f is Riemann intergrable then it is continuous on [a,b] which means that it must also be bounded? or is this incorrect i find this topic rather difficult  For part (b) i dont really have a clue i cant recall any function like the one they ask for
 A: To show that $f$ cannot be both unbounded and Riemann integrable, it is enough to produce some $\epsilon > 0$ such that for any real number $I$ and any $\delta > 0$ there is a tagged partition $P$ with $\|P\| < \delta$ and with a Riemann sum satisfying
$$|S(P,f) - I| > \epsilon$$
Since $f$ is unbounded it must be unbounded on at least one subinterval $[x_{j-1},x_j].$  Using the reverse triangle inequality we have 
$$|S(P,f) - I| = \left|f(t_j)(x_j - x_{j-1}) + \sum_{k \neq j}f(t_k)(x_k - x_{k-1}) - I \right| \\ \geqslant |f(t_j)|(x_j - x_{j-1}) - \left|\sum_{k \neq j}f(t_k)(x_k - x_{k-1}) - I \right|$$
Since $f$ is unbounded on $[x_{j-1},x_j]$, choose $t_j$ such that
$$|f(t_j)| > \frac{\epsilon  + \left|\sum_{k \neq j}f(t_k)(x_k - x_{k-1}) - I \right|}{x_j - x_{j-1}},$$
and it follows that
$$|S(f,P) - I| > \epsilon.$$
Thus, when $f$ is unbounded, it is impossible to find $I$ such that for every $\epsilon > 0$ and sufficiently fine partitions, the condition $|S(P,f) - I| < \epsilon$ holds.  We can always select the tags so that the inequality is violated.
A: (a) $1/x$ is continuous but not bounded on $(0,\infty)$, so you cannot argue boundedness from continuity. What if $x$ were unbounded in some neighborhood of the integration domain? What would that say about the upper and lower partitions?
(b) What about $x = 1$ if $x$ is rational and $0$ otherwise? Surely this function is bounded. Is it Riemann integrable?
