Prove: If $f:[a,b] \to \mathbb{R}$ is Riemann integrable, then $f$ is bounded.

My class did this using proof by contradiction and epsilon-delta definition (which I have hard time understanding).

  1. Is there an alternative proof (perhaps easier)?

  2. Can someone explain the proof by contradiction method?

I did not post the class proof because it's way too long. Thanks.

  • 1
    $\begingroup$ Hint: If $f$ is unbounded, then given a partition of $[a,b]$ and a number $M>0$, show that there are tags subordinate to the partition such that the associated Riemann sum exceeds $M$ in absolute value. $\endgroup$ – David Mitra Apr 30 '13 at 14:22
  • $\begingroup$ If $f$ were integrable with integral $L$, then given any partition with sufficiently small norm and any set of tags, the associated Riemann sum will be close to $L$. If $f$ were unbounded, then, using my previous comment, one can arrive at a contradiction (take $M$ BIG). With full rigor, you can't avoid using deltas and epsilons ... $\endgroup$ – David Mitra Apr 30 '13 at 14:33
  • $\begingroup$ @David Mitra : I like your first hint. I don't see why one needs epilsons and deltas for this problem. $\endgroup$ – Stefan Smith Apr 30 '13 at 17:28

I don't think there exists a proof that does not use the $\epsilon\mathrm{-}\delta$ definition. But the proof using that definition isn't hard. Let $T = \{t_1, t_2, \ldots, t_n\}$ and $X = \{x_0, x_1, \ldots, x_n\}$ be arbitrary points partitioning the interval $[a,b]$ so that: $$ a = x_0 \leq t_1 \leq x_1 \leq t_2 \leq \ldots \leq t_n \leq x_n = b. $$ and let $$ \Delta X = \{\Delta x_1, \Delta x_2, \ldots \Delta x_n\} = \{x_1 - x_0, x_2 - x_1, \ldots, x_n - x_{n-1}\}. $$ You are recommended to draw the points in $T$ and $X$ on a number line. Then the function is Riemann integrable on the interval $[a,b]$ iff $$ \forall \epsilon > 0, \exists \delta > 0: \mathrm{max}(\Delta X) < \delta \implies |R - I| < \epsilon, $$ where $$ R = \sum_{i=1}^nf(t_i)\Delta x_i\quad\mathrm{and}\quad I = \int_a^bf(x)dx. $$ $R$ is the Riemann sum and $I$ is the integral it approximates. The second last equation is the $\epsilon\mathrm{-}\delta$ relation; for any $\epsilon$ you give me, I can find a $\delta$ so that, if the partition of $[a,b]$ is finer than that, then the approximation of the Riemann sum will be closer to the integral than $\epsilon$.

Now we'll prove $\mathrm{RInt}(f[a,b]) \implies \mathrm{Bounded}(f[a,b])$ - if a function is Riemann integrable on an interval it must be bounded on that same interval. First some background proof rules. For all intervals we're assuming $a < b$ because proof rule (d) doesn't work otherwise:

  • a) $\mathrm{RInt}(f[a,b]) \implies (\forall \epsilon > 0, \exists \delta > 0: \mathrm{max}(\Delta X) < \delta \implies |R - I| < \epsilon)$: The $\epsilon\mathrm{-}\delta$ rule described above.
  • b) $\neg\mathrm{Bounded}(f[a,b]) \implies \exists [a', b']\subset [a,b]: \neg\mathrm{Bounded}(f[a',b'])$: If $f$ is unbounded on $[a,b]$ then there is a subinterval $[a', b']$ on which it is unbounded.
  • c) $\neg\mathrm{Bounded}(f[a,b]) \implies \sup |f[a,b]|=\infty$: The supremum of an unbounded function is positive or negative infinity.
  • d) $\sup |f[a,b]|=\infty \implies \forall c, \forall x_1 \in [a,b], \exists x_2 \in [a,b]: |f(x_2) - f(x_1)| > c$: For an unbounded function, we can always select two points in it so that the distance between them is greater than some value $c$. This is "obvious" but beyond me to actually prove.
  • e) $|a + b| \leq |a| + |b|$: Triangle inequality.

Now comes the proof by contradiction. The proof is long because I've split up the small trivial stuff over several lines - a better mathematician would write a much more succint proof:

  1. $\mathrm{RInt}(f[a,b])$: Assumption.
  2. $\neg\mathrm{Bounded}(f[a,b])$: Assumption.
  3. $\forall \epsilon > 0, \exists \delta > 0: \mathrm{max}(\Delta X) < \delta \implies |R - I| < \epsilon$: By (1) and (a).
  4. $\exists \delta > 0: \mathrm{max}(\Delta X) < \delta \implies |R - I| < 1$: By (3) with $\epsilon = 1$.
  5. $\exists \delta > 0: \mathrm{max}(\Delta X) < \delta$: Assumption.
  6. $|R - I| < 1$ Implied by (4) and (5)
  7. $\neg\mathrm{Bounded}(f[a',b'])$: By (2) and (b).
  8. $\neg\mathrm{Bounded}(f[x_{c-1}, x_c])$: By (7) with $a' = x_{c-1}$ and $b' = x_c$, where $c$ is some number $1\ldots n$, indexing the set $X$.
  9. $\sup |f[x_{c-1},x_c]| = \infty$: By (8) and (c)
  10. $\forall y, \forall x_1 \in [x_{c-1},x_c], \exists x_2 \in [x_{c-1},x_c]: |f(x_2) - f(x_1)| > y$: By (9) and (d).
  11. $|f(t'_c) - f(t_c)| > 2/\Delta x_c$: By (10) with $x_2 = t'_c$, $x_1 = f(t_c)$ and $y = 2/\Delta x_c$.
  12. $|f(t'_c) - f(t_c)|\Delta x_c > 2$: By (11).
  13. Let $T' = T - \{t_c\} + \{t'_c\}$: It's the same set as $T$ but with point $t_c$ replaced with $t'_c$.
  14. Let $R' = \sum_{i=1}^nf(t'_i)\Delta x_i$: Where $t'_i \in T'$. So it is the same Riemann sum as $R$ except with $t_c$ replaced.
  15. $|R' - I| < 1$: By (6), (14) and (a).
  16. $|R' - I| + |R - I| < 2$: By (6) and (15).
  17. $|R' - I| + |I - R| < 2$: By (16).
  18. $|R' - R| < 2$: By (17) and (e).
  19. $|\sum_{i=1}^nf(t_i)\Delta x_i - \sum_{i=1}^nf(t'_i)\Delta x_i| < 2$: By (14), (18) and the definition of the Riemann sum.
  20. $|f(t_c)\Delta x_c - f(t'_c)\Delta x_c| < 2$: By (14) and (19), remember that we only changed one term of the sum.
  21. $|f(t_c) - f(t'_c)|\Delta x_c$ < 2: By (20).
  22. $\bot 11, 20$: Line (12) and (21) contradicts each other.
  23. $\mathrm{Bounded}(f[a,b])$: By (22) and (2), a contradiction allows us to negate an assumption.
  24. $\mathrm{RInt}(f[a,b]) \implies \mathrm{Bounded}(f[a,b])$: Conclusion of (1) and (23).

Not the answer you're looking for? Browse other questions tagged or ask your own question.