Continuity of the (real) $\Gamma$ function. Consider the real valued function
$$\Gamma(x)=\int_0^{\infty}t^{x-1}e^{-t}dt$$
where the above integral means the Lebesgue integral with the Lebesgue measure in $\mathbb R$. The domain of the function is $\{x\in\mathbb R\,:\, x>0\}$, and now I'm trying to study the continuity. The function  $$t^{x-1}e^{-t}$$
is positive and bounded if $x\in[a,b]$, for $0<a<b$, so using the dominated convergence theorem in $[a,b]$, I have:
$$\lim_{x\to x_0}\Gamma(x)=\lim_{x\to x_0}\int_0^{\infty}t^{x-1}e^{-t}dt=\int_0^{\infty}\lim_{x\to x_0}t^{x-1}e^{-t}dt=\Gamma(x_0)$$
Reassuming $\Gamma$ is continuous in every interval $[a,b]$; so can I conclude that $\Gamma$ is continuous on all its domain?
 A: You could also try the basic approach by definition. 
For any $\,b>0\,\,\,,\,\,\epsilon>0\,$ choose $\,\delta>0\,$ so that $\,|x-x_0|<\delta\Longrightarrow \left|t^{x-1}-t^{x_0-1}\right|<\epsilon\,$ in $\,[0,b]\,$
:
$$\left|\Gamma(x)-\Gamma(x_0)\right|=\left|\lim_{b\to\infty}\int\limits_0^b \left(t^{x-1}-t^{x_0-1}\right)e^{-t}\,dt\right|\leq$$
$$\leq\lim_{b\to\infty}\int\limits_0^b\left|t^{x-1}-t^{x_0-1}\right|e^{-t}\,dt<\epsilon\lim_{b\to\infty}\int\limits_0^b e^{-t}\,dt=\epsilon$$
A: Since we want to consider continuity at a single fixed point $x_0$, we may assume that $x_0\in(a,b)$ for certain $0<a<1<b$. 
Assume $x\in(a,b)$. Using the Mean Value Theorem, there exists $\xi$ between $x$ and $x_0$ such that 
$$
t^{x-1}-t^{x_0-1}=(\log t)\,t^{\xi-1}\,(x-x_0).
$$
So
$$
|t^{x-1}-t^{x_0-1}|\leq|(\log t)\,(t^{1-a}+t^{b-1})|\,|x-x_0|.
$$
The estimate for $t^{\xi-1}$ is obtained by considering $t<1$ and $t>1$ respectively. 
Now
\begin{align}
\left|\Gamma(x)-\Gamma(x_0)\right|
&\leq\int_0^\infty |t^{x-1}-t^{x_0-1}|\,e^{-t}\,dt\\ \ \\
&\leq|x-x_0|\,\int_0^\infty|(\log t)\,(t^{1-a}+t^{b-1})|\,e^{-t}\,dt\tag{1}\\ \ \\
&\leq\,|x-x_0|\,\left(\frac1{(2-a)^2}+\frac1{b^2}+\frac{2c}{\sqrt e}\right)
\end{align}
(see below), and so $\Gamma$ is continuous at $x_0$. Note that $(1)$ is a Lipschitz condition, but it is local: in the sense that the constant depends on $a$ and $b$. 

Note that 
$$\tag{2}
\left|\int_0^1 t^r\,\log t\,dt\right|<\infty \iff r>-1.
$$
Since $0<a<1$ and $b>0$, we have $1-a>0$ and $b-1>-1$. By $(2)$, then 
$$\tag{3}
\int_0^1|(\log t)\,(t^{1-a}+t^{b-1})|\,e^{-t}\,dt\leq-\int_0^1\log t\,(t^{1-a}+t^{b-1})\,dt=\frac1{(2-a)^2}+\frac1{b^2}
$$
Let $$c=\sup\{|\log t\,(t^{1-a}+t^{b-1})e^{-t/2}|:\ t\in(0,\infty)\}$$ (the sup exists because the function is continuous and bounded, since both the limits at $0$ and $\infty$ exist). Then
\begin{align}\tag{4}
\int_1^\infty|(\log t)\,(t^{1-a}+t^{b-1})|\,e^{-t}\,dt
&=\int_1^\infty|(\log t)\,(t^{1-a}+t^{b-1})|\,e^{-t/2}\,e^{-t/2}\,dt\\ \ \\
&\leq c\,\int_1^\infty e^{-t/2}\,dt=\frac{2c}{\sqrt e}.
\end{align}
Using $(3)$ and $(4)$, we get
$$
\int_0^\infty|(\log t)\,(t^{1-a}+t^{b-1})|\,e^{-t}\,dt\leq\frac1{(2-a)^2}+\frac1{b^2}+\frac{2c}{\sqrt e}.
$$
A: Note that it suffices to show the continuity of $\Gamma(x)$ for all $x>1$ because we can use the known property $\Gamma(x+1)=x\Gamma(x)$ to conclude continuity of those $x$ with $0<x\leq1$. I will break down the proof into  two steps:
$1.)$ I show that $\Gamma_k(x):=\int\limits_0^k t^{x-1}e^{-t}dt$ is continuous at point $a>1$.
$2.)$ I show that $\Gamma_k(x)\to\Gamma(x)$ uniformly on a neighbourhood of $a$ so that the limit function inherits the continuity on this neighbourhood.

Proof:
$1.)$
The integrand $K:(1,\infty)\times(0,k)$, where $K(x,t)=t^{x-1}e^{-t}$ is continuous in each point. For a given $\epsilon>0$ and a point $(a,t')$, there exists a $\delta_{t'}$ such that
$$\Big\Vert{x\choose t}-{a\choose t'}\Big\Vert<\delta_{t'}\implies | K(x,t)-K(a,t'|<\frac{\epsilon}{2k}.$$ We define a neigbourhood of $t'$ by
$$U_{\delta_t'}(t'):=\{t\in(0,k)\mid|t-t'|<\delta_{t'}\}.$$
The union $\bigcup\limits_{t\in(0,k)}U_{\delta_t}(t)$ covers the interval $(0,k)$ and due to compactness there exists a finite cover $\bigcup\limits_{i=1}^nU_{\delta_{t_i}}(t_i)$ of $(0,k)$. Now we set $\delta:=\min\{\delta_1\cdots, \delta_n\}$. Hence, for all $x>1$ with $|x-a|<\delta$ it follows:
$$
|\Gamma_k(x)-\Gamma_k(a)|=\left|\int\limits_0^k K(x,t)dt-\int\limits_0^k K(a,t)dt\right|\leq\int\limits_0^k \left|K(x,t)-K(a,t)\right|dt\\
\leq\int\limits_0^k \left|K(x,t)-K(a,t_i)|+|K(a,t_i)-K(a,t)\right|dt<\int\limits_0^k\frac{\epsilon}{k} dt<\epsilon.
$$
Here we used that $K$ is continuous in each $(a,t_i)$ and that $t$ lies witihin a  $U_{\delta_{t_i}}(t_i)$ so that $\Vert{x\choose t}-{a\choose t_i}\Vert<\delta_i$ and that $\Vert{a\choose t_i}-{a\choose t_i}\Vert<\delta_i$. Hence, $\Gamma_k(x)$ is continuous.
$2.)$ Now let's consider only those $x$ with $1<x<a+1$. Then, we see that
$$
\left|\Gamma_k(x)-\Gamma(x)\right|\leq \int\limits_k^{\infty}\left| t^{x-1}e^{-t}dt\right|dt\leq\int\limits_k^{\infty}\left| t^{a}e^{-t}dt\right|dt=\Gamma_k(a+1)-\Gamma(a+1)<\epsilon
$$
as $\lim\limits_{k\to\infty}\Gamma_k(a)=\Gamma(a)$. Hence, we have uniform convergence on the neighbourhood of $a$ and $\Gamma(x)$ inherits the continuity at point $a$.
