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?

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
$$
\int_0^\infty|(\log t)\,(t^{1-a}+t^{b-1})|\,e^{-t}\,dt\leq\frac1{(2-a)^2}+\frac1{b^2}+\frac{2c}{\sqrt e}.
$$
