Gamma Function is Continuous on Positive Reals

Theorem
Let $\Gamma: \R_{>0} \to \R$ be the Gamma function, restricted to the strictly positive real numbers.

Then $\Gamma$ is continuous.

Proof
Let $0 < \alpha < a \le x \le y \le b < \beta$.

Let $0 < \delta < \Delta$.

Then:


 * $(1): \quad \ds \size {\int_\delta^\Delta t^{x - 1} e^{-t} \rd t - \int_\delta^\Delta t^{y - 1} e^{-t} } \le \int_\delta^\Delta \paren {t^{x - 1} - t^{y - 1} }e^{-t} \rd t$

From the Mean Value Theorem:
 * $(2): \quad \dfrac {t^{x - 1} - t^{y - 1} }{x - y} = \paren {\ln t} t^{\xi - 1}$

for some $\xi \in \R$ such that $x \le \xi \le y$.

Let $r \in \R_{>0}$.

Then:
 * $t^{-r} \ln t \to 0$ as $t \to \infty$
 * $t^r \ln t \to 0$ as $t \to 0^+$

From $(2)$ it follows that:
 * $\exists H \in \R: \size {t^{x - 1} - t^{y - 1} } \le H \paren {t^{\alpha - 1} + t^{\beta - 1} } \size {x - y}$

From $(1)$ it follows that:
 * $\size {\map \Gamma x - \map \Gamma y} \le H \size {x - y} \paren {\map \Gamma \alpha - \map \Gamma \beta}$

The result follows from the Squeeze Theorem.