Monotone Function is Riemann Integrable

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Theorem

Let $\closedint a b$ be a closed real interval, where $a < b$.

Let $f: \closedint a b \to \R$ be a monotone real function.


Then $f$ is Riemann integrable over $\closedint a b$.


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:

$n > \dfrac {\paren {b - a} \paren {\map f b - \map f a} } {\epsilon}$

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

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

Then:

$a = x_0 < x_1 < \cdots < x_n = b$

and hence $P = \set {x_0, x_1, \ldots, x_n}$ is a subdivision of $\closedint a b$.


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

$\displaystyle \map L P = \sum_{k \mathop = 1}^n \map f {x_{k - 1} } \paren {x_k - x_{k - 1} } = \frac {b - a} n \sum_{k \mathop = 1}^n \map f {x_{k - 1} }$
$\displaystyle \map U P = \sum_{k \mathop = 1}^n \map f {x_k} \paren {x_k - x_{k - 1} } = \frac {b - a} n \sum_{k \mathop = 1}^n \map f {x_k}$

Therefore:

\(\displaystyle 0\) \(\le\) \(\displaystyle \overline {\int_a^b} \map f x \rd x - \underline {\int_a^b} \map f x \rd x\)
\(\displaystyle \) \(\le\) \(\displaystyle \map U P - \map L P\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {b - a} \paren {\map f b - \map f a} } n\)
\(\displaystyle \) \(<\) \(\displaystyle \epsilon\)

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

$\blacksquare$


Also see


Sources