Integrability Theorem for Functions Continuous on Open Intervals

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

If $f$ is continuous on $\left({a \,.\,.\, b}\right)$ and $\displaystyle \lim_{x \to a^+} f\left({x}\right)$ and $\displaystyle \lim_{x \to b^-} f\left({x}\right)$ exist, then $f$ is Riemann integrable on $\left[{a \,.\,.\, b}\right]$.

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


 * $U(S)–L(S)<\epsilon$

where $U(S)$ and $L(S)$ are respectively the upper and lower sums of $f(x)$ 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 maximum of this bound and $\max \left({\left\vert{f \left(a\right)}\right\vert, \left\vert{f \left(b\right)}\right\vert}\right)$ serves as a bound for $f$ throughout its domain $\left[{a \,.\,.\, b}\right]$.

Therefore, $f$ is bounded.

Since $f$ is bounded, a positive bound $K$ exists for $f$ on $\left[{a \,.\,.\, b}\right]$.

Let a positive $\epsilon$ be given, and choose a $\delta$ that satisfies:


 * $0 < \delta < min(\epsilon/(3*2*K), (b-a)/2)$

Since $f$ is continuous on $\left({a \,.\,.\, b}\right)$, $f$ is continuous on $[a+\delta..b-\delta]$ because this interval is a subset of $\left({a \,.\,.\, b}\right)$ as $\delta$ > 0.

By Continuous Function is Riemann Integrable, $f$ is integrable on $[a+\delta..b-\delta]$.

Since $f$ is integrable on $[a+\delta..b-\delta]$, there exists a subdivision $S_\delta$ of $[a+\delta..b-\delta]$ that satisfies:


 * $U(S_\delta)–L(S_\delta)<\epsilon/3$

where $U(S_\delta)$ and $L(S_\delta)$ are, respectively, the upper and lower sums of $f$ on $[a+\delta..b-\delta]$ with respect to the subdivision $S_\delta$.

Define the following subdivision of $\left[{a \,.\,.\, b}\right]$: $S = S_\delta \cup \left\{{a,b}\right\}$.

The upper sum of $f(x)$ on $\left[{a \,.\,.\, b}\right]$ with respect to $S$ is per definition:


 * $U(S) = M_a*\delta + U(S_\delta) + M_b*\delta$

where $M_a$ is the supremum of $f$ on $[a..a+\delta]$, and $M_b$ is the supremum of $f$ on $[b-\delta..b]$.

$M_a$ and $M_b$ exist by the least upper bound property of the real numbers because $f$ is bounded on $[a..a+\delta]$ and $[b-\delta..b]$.

The lower sum of $f(x)$ on $\left[{a \,.\,.\, b}\right]$ with respect to $S$ is per definition:


 * $L(S) = m_a*\delta + L(S_\delta) + m_b*\delta$

where $m_a$ is the infimum of $f$ on $[a..a+\delta]$, and $m_b$ is the infimum of $f$ on $[b-\delta..b]$.

$m_a$ and $m_b$ exist by the greatest lower bound property of the real numbers because $f$ is bounded on $[a..a+\delta]$ and $[b-\delta..b]$.

Define the sum

Define the sum

Therefore, $U'$ and $L'$ satisfy:


 * $U' \geq U(S)$


 * $L' \leq L(S)$

From these two inequalities follows:

Hence


 * $U(S)–L(S)< \epsilon$