Monotone Real Function is Darboux Integrable
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 Darboux 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 Axiom of Archimedes, 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 Darboux sum and upper Darboux sum of $f$ belonging to the subdivision $P$ are, respectively:
| \(\ds \map L P\) | \(=\) | \(\, \ds \sum_{k \mathop = 1}^n \map f {x_{k - 1} } \paren {x_k - x_{k - 1} } \, \) | \(\, \ds = \, \) | \(\ds \frac {b - a} n \sum_{k \mathop = 1}^n \map f {x_{k - 1} }\) | ||||||||||
| \(\ds \map U P\) | \(=\) | \(\, \ds \sum_{k \mathop = 1}^n \map f {x_k} \paren {x_k - x_{k - 1} } \, \) | \(\, \ds = \, \) | \(\ds \frac {b - a} n \sum_{k \mathop = 1}^n \map f {x_k}\) |
Therefore:
| \(\ds 0\) | \(\le\) | \(\ds \overline {\int_a^b} \map f x \rd x - \underline {\int_a^b} \map f x \rd x\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds \map U P - \map L P\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\paren {b - a} \paren {\map f b - \map f a} } n\) | ||||||||||||
| \(\ds \) | \(<\) | \(\ds \epsilon\) |
Since $\epsilon \in \R_{>0}$ is arbitrary, the result follows.
$\blacksquare$
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Also see
Sources
- 1953: Walter Rudin: Principles of Mathematical Analysis ... (previous): $6.9$