Continuous Real Function is Darboux Integrable

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

Then $f$ is always Riemann integrable.

Proof
As $f$ is continuous on $\left[{a \,. \, . \, b}\right]$, it follows from the Continuity Property that $f$ is bounded on $\left[{a \,. \, . \, b}\right]$.

Let $P = \left\{{x_0, x_1, x_2, \ldots, x_{n-1}, x_n}\right\}$ be a subdivision of $\left[{a \,. \, . \, b}\right]$.

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

That is:
 * $\displaystyle L \left({P}\right) = \sum_{k=1}^n m_{k} \left({x_{k} - x_{k - 1}}\right)$

where
 * $\displaystyle m_k = \inf_{x \in \left[{x_{k - 1} \, . \, . \, x_{k}}\right]} f \left({x}\right)$

Let $f$ be bounded above on $\left[{a \,. \, . \, b}\right]$ by $H$.

Thus $\forall t \in \left[{a \,. \, . \, b}\right]: f \left({t}\right) \le H$.

In particular, $\forall 0 \le k \le n: m_k \le H$. Then:

Thus $L \left({P}\right)$ is bounded above by $H \left({b - a}\right)$.

Hence it has a supremum and we can define $\displaystyle \int_a^b f \left({x}\right) dx = \sup_P L \left({P}\right)$, where $\sup_P$ extends over all possible partitions of $\left[{a \,. \, . \, b}\right]$.

In the same way, we can consider the upper sum $U \left({P}\right)$ of $f \left({x}\right)$ on $\left[{a \,. \, . \, b}\right]$ belonging to the subdivision $P$.

Similarly, we can show that $U \left({P}\right)$ is bounded below and has an infimum.

After a little algebra, it is apparent that the use of the definition based on the upper sum would yield the result $\displaystyle - \int_a^b -f \left({x}\right) dx = \inf_P \left({U \left({P}\right)}\right)$.

Now, if $f$ is continuous on $\left[{a \,. \, . \, b}\right]$, it follows from basic results in differentiation that $F$ is a primitive for $f$ on $\left[{a \,. \, . \, b}\right]$ iff $-F$ is a primitive for $-f$.

The result follows.