# Continuous Function is Riemann Integrable

## Theorem

Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.

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

## 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 $\closedint a b$ such that:

$\map U P – \map L P < \epsilon$

where $\map U P$ and $\map L P$ denote the upper sum and lower sum of $\map f x$ on $\closedint a b$ 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 $x, y \in \closedint a b$ are such that $\size {x – y} < \delta$, then:

$\size {\map f x – \map f y} < \dfrac \epsilon {b - a}$

Let $P = \set {x_0, x_1, x_2, \ldots, x_n}$ be a subdivision of $\closedint a b$ such that:

$\displaystyle \max_{1 \mathop \le k \mathop \le n} \paren {x_k – x_{k - 1} } < \delta$

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

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

 $\displaystyle \map f {u_k}$ $=$ $\displaystyle \sup \set {\map f x: x \in \closedint {x_{k - 1} } {x_k} }$ $\displaystyle \map f {v_k}$ $=$ $\displaystyle \inf \set {\map f x: x \in \closedint {x_{k - 1} } {x_k} }$

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

$\size {u_k – v_k} < \delta$

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

$\displaystyle \map f {u_k} – \map f {v_k} < \frac \epsilon {b – a}$

This gives:

 $\displaystyle \map U P – \map L P$ $=$ $\displaystyle \sum_{k \mathop = 1}^n \map f {u_k} \paren {x_k – x_{k - 1} } - \sum_{k \mathop = 1}^n \map f {v_k} \paren {x_k – x_{k - 1} }$ $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 1}^n \paren {\map f {u_k} – \map f {v_k} } \paren {x_k – x_{k - 1} }$ $\displaystyle$ $<$ $\displaystyle \frac \epsilon {b – a} \sum_{k \mathop = 1}^n \paren {x_k – x_{k - 1} }$ $\displaystyle$ $=$ $\displaystyle \frac \epsilon {b – a} \paren {x_n – x_0}$ $\displaystyle$ $=$ $\displaystyle \epsilon$

as desired.

$\blacksquare$

## Notes

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