Riemann integral

Setup: partitions

Let $f : [a, b] \to R$ be a bounded function.

A partition of $[a, b]$ is a finite ordered set

P = (x_{0}, \dots, x_{n}), a = x_{0} < x_{1} < \dots < x_{n} = b .

We denote $P [a, b]$ , the set of all partitions of $[a, b]$ .
The norm of $P$ is

| P | := max_{1 \leq i \leq n} (x_{i} - x_{i - 1}) .

We say $P^{'}$ is a refinement of $P$ if it contains all points of $P$ (and possibly more).

Lower/upper (Darboux) sums

For each subinterval $[x_{i - 1}, x_{i}]$ define

m_{i} := inf_{x \in [x_{i - 1}, x_{i}]} f (x), M_{i} := sup_{x \in [x_{i - 1}, x_{i}]} f (x) .

Then the lower sum and upper sum are

m (f, P) := \sum_{i = 1}^{n} m_{i} (x_{i} - x_{i - 1}), M (f, P) := \sum_{i = 1}^{n} M_{i} (x_{i} - x_{i - 1}) .

Riemann sums

A Riemann sum chooses arbitrary points $ξ_{i} \in [x_{i - 1}, x_{i}]$ and defines

S (f, P, ξ) := \sum_{i = 1}^{n} f (ξ_{i}) (x_{i} - x_{i - 1}) .

Always $m (f, P) \leq S (f, P, ξ) \leq M (f, P)$ .

Integrability

Define the lower integral and upper integral of $f$ by

\underset{―}{\int_{a}^{b}} f := sup_{P} m (f, P), \overset{―}{\int_{a}^{b}} f := inf_{P} M (f, P),

where $P$ ranges over all partitions of $[a, b]$ , i.e, $P \in P [a, b]$ .

We say $f$ is Riemann integrable on $[a, b]$ if

\underset{―}{\int_{a}^{b}} f = \overset{―}{\int_{a}^{b}} f .

In that case, their common value is denoted by

\int_{a}^{b} f (x) d x .

Equivalent practical criterion: $f$ is Riemann integrable iff for every $ε > 0$ there exists a partition $P$ such that

M (f, P) - m (f, P) < ε .

Convergence of Riemann sums

If $f$ is Riemann integrable and $(P_{n})$ is any sequence of partitions with $| P_{n} | \to 0$ , then for any choice of tags $ξ_{i}^{(n)} \in [x_{i - 1}^{(n)}, x_{i}^{(n)}]$ one has

\int_{a}^{b} f (x) d x = lim_{n \to \infty} S (f, P_{n}, ξ^{(n)}) .

Standard non-example (Dirichlet function)

Let $ψ : [0, 1] \to R$ be

ψ (x) = {\begin{cases} 1, & x \in Q, \\ 0, & x \notin Q . \end{cases}

On every subinterval there are rationals and irrationals, hence every partition $P$ satisfies $m (ψ, P) = 0$ and $M (ψ, P) = 1$ . Therefore

\underset{―}{\int_{0}^{1}} ψ = 0 \neq 1 = \overset{―}{\int_{0}^{1}} ψ,

so $ψ$ is not Riemann integrable.