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:

$(1): \quad \displaystyle \left|{\int_\delta^\Delta t^{x - 1} e^{-t} \ \mathrm d t - \int_\delta^\Delta t^{y - 1} e^{-t} }\right| \le \int_\delta^\Delta \left({t^{x - 1} - t^{y - 1}}\right)e^{-t} \ \mathrm d t$

From the Mean Value Theorem:

$(2): \quad \dfrac {t^{x - 1} - t^{y - 1} }{x - y} = \left({\ln t}\right) t^{\xi - 1}$

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:

$\exists H \in \R: \left|{t^{x - 1} - t^{y - 1} }\right| \le H \left({t^{\alpha - 1} + t^{\beta - 1} }\right) \left|{x - y}\right|$

From $(1)$ it follows that:

$\left|{\Gamma \left({x}\right) - \Gamma \left({y}\right) }\right| \le H \left|{x - y}\right| \left({\Gamma \left({\alpha}\right) - \Gamma \left({\beta}\right) }\right)$

The result follows from the Squeeze Theorem.

$\blacksquare$