Continuous Function is Riemann Integrable

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Theorem

Let $f$ be a real function which is continuous on the closed interval $\left[{a \,.\,.\, b}\right]$.

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


Proof

We have the result $f$ is bounded by Continuous Real Function is Bounded.

By Condition for Riemann Integrability, it suffices to show that for all $\epsilon > 0$, there exists a subdivision $P$ of $\left[{a \,.\,.\, b}\right]$ such that:

$U \left({P}\right) – L \left({P}\right) < \epsilon$

where $U \left({P}\right)$ and $L \left({P}\right)$ denote the upper sum and lower sum of $f \left({x}\right)$ on $\left[{a \,.\,.\, b}\right]$ belonging to the subdivision $P$.

Let $\epsilon > 0$.

We have the result Continuous Function on Closed Interval is Uniformly Continuous.

By the definition of uniform continuity, there exists a $\delta > 0$ such that if $x, y \in \left[{a \,.\,.\, b}\right]$ are such that $\left\vert{x – y}\right\vert < \delta$, then:

$\left\vert{f \left({x}\right) – f \left({y}\right)}\right\vert < \dfrac \epsilon {b - a}$

Let $P = \left\{ {x_0, x_1, x_2, \ldots, x_n} \right\}$ be a subdivision of $\left[{a \,.\,.\, b}\right]$ such that:

$\displaystyle \max_{1 \mathop \le k \mathop \le n} \left({x_k – x_{k - 1} }\right) < \delta$

For all integers $k$ satisfying $1 \le k \le n$, it follows from the Heine-Borel theorem that $\left[{x_{k - 1} \,.\,.\, x_k}\right]$ is compact.

So we can apply Corollary 3 to Continuous Image of Compact Space is Compact to conclude that there exist $u_k, v_k \in \left[{x_{k - 1} \,.\,.\, x_k}\right]$ such that:

$\displaystyle f \left({u_k}\right) = \sup \left\{ {f \left({x}\right): x \in \left[{ x_{k - 1} \,.\,.\, x_k}\right]} \right\}$
$\displaystyle f \left({v_k}\right) = \inf \left\{ {f \left({x}\right): x \in \left[{ x_{k - 1} \,.\,.\, x_k}\right]} \right\}$

By assumption, $x_k – x_{k - 1} < \delta$, so:

$\left\vert{u_k – v_k}\right\vert < \delta$

It follows from the definition of $\delta$ that:

$\displaystyle f \left({u_k}\right) – f \left({v_k}\right) < \frac \epsilon {b – a}$

This gives:

\(\displaystyle U \left({P}\right) – L \left({P}\right)\) \(=\) \(\displaystyle \sum_{k \mathop = 1}^n f \left({u_k}\right) \left({ x_k – x_{k-1} }\right) - \sum_{k \mathop = 1}^n f \left({v_k}\right) \left({ x_k – x_{k-1} }\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 1}^n \left[{ f \left({u_k}\right) – f \left({v_k}\right) }\right] \left({ x_k – x_{k-1} }\right)\) $\quad$ $\quad$
\(\displaystyle \) \(<\) \(\displaystyle \frac \epsilon {b – a} \sum_{k \mathop = 1}^n \left({ x_k – x_{k-1} }\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac \epsilon {b – a} \left({x_n – x_0}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \epsilon\) $\quad$ $\quad$

as desired.

$\blacksquare$


Sources


Notes

If $f$ is not continuous on $\left[{a \,.\,.\, b}\right]$ (while still bounded), then the two definitions do not necessarily have the same result. But they might.