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
It suffices to show that, for a given strictly positive $\epsilon$, a subdivision $S$ of $\left[{a \,.\,.\, b}\right]$ exists such that


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

where $U\left({S}\right)$ and $L\left({S}\right)$ are respectively the upper and lower sums of $f$ on $\left[{a \,.\,.\, b}\right]$ with respect to the subdivision $S$.

We start by showing that $f$ is bounded.

Consider the restriction of $f$ to $\left({a \,.\,.\, b}\right)$.

By Extendability Theorem for Functions Continuous on Open Intervals there exists a continuous function $g$ that is defined on $\left[{a \,.\,.\, b}\right]$, and that equals $f$ on $\left({a \,.\,.\, b}\right)$.

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

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

Since $g$ is bounded on $\left({a \,.\,.\, b}\right)$, so is $f$, because $f=g$ on $\left({a \,.\,.\, b}\right)$.

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

Since $f$ is bounded on $\left({a \,.\,.\, b}\right)$, a bound exists for $f$ on $\left({a \,.\,.\, b}\right)$.

The greatest of this bound and the maximum $\max \left({\left\vert{f \left({a}\right)}\right\vert, \left\vert{f \left({b}\right)}\right\vert}\right)$ serves as a bound for $f$ throughout its domain $\left[{a \,.\,.\, b}\right]$.

Therefore, $f$ is bounded.

By Bounded Function Continuous on Open Interval is Riemann Integrable, $f$ is Riemann integrable.