# 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

• Traditionally Riemann integrable doesn't imply continuous (the function $f(x) = x$ everywhere except at $x = 2$, where $f(2) = 5$ is generally considered to be Riemann integrable, but not continuous). It does imply almost everywhere continuity though. – Mark Apr 25 '17 at 1:23
• Note that the presentation of Riemann integral using Darboux sums is not possible for unbounded functions because these sums are defined in terms of bounds of the function. Also as RRL's excellent answer shows, the Riemann's definition also cannot be used if the function is unbounded, it makes perfect sense to define the concept of Riemann integration only for bounded functions. And if needed use improper integrals to handle unbounded functions/intervals. – Paramanand Singh Apr 25 '17 at 3:21

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.

• Perfect reason for restricting the definition of Riemann integral to bounded functions only. +1 – Paramanand Singh Apr 25 '17 at 3:17

(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?

• $\frac{1}{x}$ is bounded on closed intervals where it is defined. – dannum Apr 25 '17 at 1:21
• and $(0,\infty)$ is an open interval. – Mortified Through Math Apr 25 '17 at 1:22
• The question says $f: [a,b] \to \mathbb R$ – dannum Apr 25 '17 at 1:23
• To see that $1/x$ is not bounded on $(0,\infty)$, choose some real $a>0$. If $1/x$ were bounded by $a$ then $1/x < a$ or $x > 1/a$. But since $a$ is nonzero $1/(a+1)$ is in $(0,\infty)$ for every $a$. – Mortified Through Math Apr 25 '17 at 1:26
• The point still stands. Thinking boundedness follows from continuity is a mistake I see often. – Mortified Through Math Apr 25 '17 at 1:27