Integrability Theorem for Functions Continuous on Open Intervals

Theorem
Let $f$ be a real function defined on an interval $\left[{a \,.\,.\, b}\right]$ such that $a < b$.

Let $f$ be continuous on $\left({a \,.\,.\, b}\right)$.

Let the one-sided limits $\displaystyle \lim_{x \mathop \to a^+} f \left({x}\right)$ and $\displaystyle \lim_{x \mathop \to b^-} f \left({x}\right)$ exist.

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

Proof
We start by showing that $f$ is bounded.

By Extendability Theorem for Functions Continuous on Open Intervals there exists a function $g$ defined on $\left[{a \,.\,.\, b}\right]$ such that:


 * $g$ equals $f$ on $\left({a \,.\,.\, b}\right)$
 * $g$ is continuous on $\left[{a \,.\,.\, b}\right]$

Therefore:


 * $g$ is bounded on $\left[{a \,.\,.\, b}\right]$ by Continuous Function on Compact Space is Bounded and Closed Real Interval is Compact

Accordingly, $g$ is bounded on $\left({a \,.\,.\, b}\right)$ as $\left({a \,.\,.\, b}\right)$ is a subset of $\left[{a \,.\,.\, b}\right]$.

Therefore, $f$ is bounded on $\left({a \,.\,.\, b}\right)$ as $f$ equals $g$ on $\left({a \,.\,.\, b}\right)$.

Also, $f$ is bounded on the set $\left\{{a, b}\right\}$ by the maximum $\max \left({\left\vert{f \left({a}\right)}\right\vert, \left\vert{f \left({b}\right)}\right\vert}\right)$.

Therefore:


 * $f$ is bounded on $\left[{a \,.\,.\, b}\right]$

We have:


 * $f$ is bounded on $\left[{a \,.\,.\, b}\right]$
 * $f$ is continuous on $\left({a \,.\,.\, b}\right)$

Therefore:


 * $f$ is Riemann integrable on $\left[{a \,.\,.\, b}\right]$ by Bounded Function Continuous on Open Interval is Riemann Integrable