The
definite integral
is usually introduced as a limit of sums. To make that rigorous — and to say which functions
have an integral at all — we squeeze the area between two approximations that never lie about it: one
that always undershoots and one that always overshoots.
Chop [a, b] with a partition
P = \{a = x_0 < x_1 < \dots < x_n = b\}, with widths
\Delta x_i = x_i - x_{i-1}. On each subinterval let
m_i = \inf_{[x_{i-1}, x_i]} f, \qquad M_i = \sup_{[x_{i-1}, x_i]} f.
The lower sum stacks rectangles at the floor of f; the
upper sum stacks them at the ceiling:
L(f, P) = \sum_{i=1}^{n} m_i\,\Delta x_i, \qquad U(f, P) = \sum_{i=1}^{n} M_i\,\Delta x_i.
Since m_i \le M_i, always L(f, P) \le U(f, P):
the true area is trapped between them.
Centrepiece: refinement, the criterion, and why continuity suffices
Step A — refining a partition raises L and lowers U
Add one new point c inside a subinterval
[x_{i-1}, x_i], splitting it into a left piece
[x_{i-1}, c] and a right piece [c, x_i].
A1 — the infimum can only go up on a smaller set. The inf over a sub-piece is
\ge the inf over the whole, so writing m',
m'' for the two pieces' infima, m' \ge m_i and
m'' \ge m_i. Hence the lower sum's contribution can only increase:
m_i\,\Delta x_i = m_i(c - x_{i-1}) + m_i(x_i - c) \le m'(c - x_{i-1}) + m''(x_i - c).
A2 — symmetrically, the supremum can only go down, so the upper sum's contribution can
only decrease. Refining therefore gives
L(f, P) \le L(f, P') \quad\text{and}\quad U(f, P') \le U(f, P) \qquad (P' \text{ a refinement of } P).
A3 — every lower sum is below every upper sum. Given any two partitions
P_1, P_2, refine to their common refinement
P_1 \cup P_2 and chain the inequalities:
L(f, P_1) \le L(f, P_1 \cup P_2) \le U(f, P_1 \cup P_2) \le U(f, P_2).
So the set of lower sums sits entirely below the set of upper sums. The two then have a clean gate
between them.
Step B — the definition: \sup L = \inf U
Define the lower integral and upper integral by taking the best
approximation from each side:
\underline{\int_a^b} f = \sup_P L(f, P), \qquad \overline{\int_a^b} f = \inf_P U(f, P).
By Step A3 always \underline{\int} f \le \overline{\int} f. We call
f Riemann integrable exactly when these agree, and then
their common value is the integral:
\underline{\int_a^b} f = \overline{\int_a^b} f =: \int_a^b f.
Cauchy criterion (the practical test). Squeezing the two integrals together is the
same as making a single upper–lower gap small. f is integrable iff for every
\varepsilon > 0 there is a partition P with
U(f, P) - L(f, P) = \sum_{i=1}^{n} (M_i - m_i)\,\Delta x_i < \varepsilon.
Step C — a continuous f on [a,b] is integrable
C1 — upgrade continuity to uniform continuity. By
Heine–Cantor,
a function continuous on the closed bounded interval [a, b] is
uniformly continuous: given \varepsilon > 0, there is one
\delta > 0 with
|x - y| < \delta \Rightarrow |f(x) - f(y)| < \dfrac{\varepsilon}{b - a}.
C2 — pick a fine enough partition. Choose P with every
\Delta x_i < \delta. On each subinterval the oscillation is small:
because f is continuous it attains its max and min there (EVT), at points no
more than \delta apart, so
M_i - m_i = f(\text{argmax}) - f(\text{argmin}) < \frac{\varepsilon}{b - a}.
C3 — assemble the Cauchy gap. Sum the oscillations against the widths:
U(f, P) - L(f, P) = \sum_i (M_i - m_i)\,\Delta x_i < \frac{\varepsilon}{b - a} \sum_i \Delta x_i = \frac{\varepsilon}{b - a}\,(b - a) = \varepsilon.
The Cauchy criterion is met for every \varepsilon, so a continuous function
on [a, b] is Riemann integrable. Uniform continuity is precisely the lever:
it makes the oscillations M_i - m_i small everywhere at once.
Let f be bounded on [a, b]. Then:
-
For any partitions, L(f, P_1) \le U(f, P_2); refinement raises
L and lowers U.
-
f is Riemann integrable iff
\displaystyle\sup_P L(f,P) = \inf_P U(f,P), equivalently iff for every
\varepsilon > 0 some P has
U(f,P) - L(f,P) < \varepsilon (the Cauchy criterion).
-
Every function continuous on [a, b] is Riemann
integrable (uniform continuity makes the oscillations M_i - m_i uniformly
small). So is every monotone function.
Riemann integration has a hard limit. Consider the indicator of the rationals on
[0, 1],
\mathbf{1}_{\mathbb{Q}}(x) = \begin{cases} 1, & x \in \mathbb{Q}, \\ 0, & x \notin \mathbb{Q}. \end{cases}
Every subinterval, however tiny, contains both a rational and an irrational (both are dense). So on
each piece m_i = 0 and M_i = 1, giving
L(f, P) = 0 \;\text{for all } P, \qquad U(f, P) = 1 \;\text{for all } P,
hence \underline{\int} = 0 \ne 1 = \overline{\int}: the lower and upper
integrals never meet, and \mathbf{1}_{\mathbb{Q}} is not Riemann
integrable. The fix is to stop chopping the domain into intervals and instead chop
the range into level sets, measuring how big each level set is — that is the
Lebesgue integral,
for which \int_0^1 \mathbf{1}_{\mathbb{Q}} = 0 because the rationals have
measure zero.