Monotone Function is Riemann Integrable

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Theorem

Let $\left[{a \,.\,.\, b}\right]$ be a closed real interval, where $a < b$.

Let $f: \left[{a \,.\,.\, b}\right] \to \R$ be a monotone real function.


Then $f$ is Riemann integrable over $\left[{a \,.\,.\, b}\right]$.


Proof

We consider the case where $f$ is increasing; the case where $f$ is decreasing is handled similarly.


Let $\epsilon \in \R_{>0}$ be a strictly positive real number.

By the Archimedean Principle, there exists a natural number $n$ such that:

$\displaystyle n > \frac {\left({b - a}\right) \left({f \left({b}\right) - f \left({a}\right)}\right)} {\epsilon}$

For $k \in \left\{{0, 1, 2, \ldots, n}\right\}$, define:

$\displaystyle x_k = a + k \frac {b - a} n$

Then $a = x_0 < x_1 < \cdots < x_n = b$, and hence $P = \left\{{x_0, x_1, \ldots, x_n}\right\}$ is a subdivision of $\left[{a \,.\,.\, b}\right]$.


The lower sum and upper sum of $f$ belonging to the subdivision $P$ are, respectively:

$\displaystyle L \left({P}\right) = \sum_{k \mathop = 1}^n f \left({x_{k-1}}\right) \left({x_k - x_{k-1}}\right) = \frac {b - a} n \sum_{k \mathop = 1}^n f \left({x_{k-1}}\right)$
$\displaystyle U \left({P}\right) = \sum_{k \mathop = 1}^n f \left({x_k}\right) \left({x_k - x_{k-1}}\right) = \frac {b - a} n \sum_{k \mathop = 1}^n f \left({x_k}\right)$

Therefore:

$\displaystyle 0 \le \overline{\int_a^b} f \left({x}\right) \ \mathrm d x - \underline{\int_a^b} f \left({x}\right) \ \mathrm d x \le U \left({P}\right) - L \left({P}\right) = \frac {\left({b - a}\right) \left({f \left({b}\right) - f \left({a}\right)}\right)} n < \epsilon$

Since $\epsilon \in \R_{>0}$ is arbitrary, the result follows.

$\blacksquare$


Also see


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