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