Theorem. Let $f_{n}:G\to \mathbf{C}, n\in \mathbf{N}$, continuous and integrable and let $f=\lim_{n\to \infty}f_{n}$ be locally uniform in $G$. Then $f$ is continuous and integrable in $G$.
Proof (old version, new is below)
Assume integrable to mean Riemann integrable. Then let $ \begin{align*} T &:= \inf \sum_{k=1}^{n}(x_{k}-x_{k-1})\cdot \sup_{x_k-1 \lt x\lt x_{k}} f(x), \\ t &:= \sup \sum_{k=1}^{n}(x_{k}-x_{k-1})\cdot \inf_{x_k-1\lt x \lt x_{k}} f(x), \\ \epsilon_{n} & = \sup_{x\in G}{\ | f_{n}(x)-f(x)|} \end{align*} $
so
$f_{n}(x) - \epsilon_{n} \le f(x) \le f_{n}(x) + \epsilon_{n} \ \forall x\in G, n\in \mathbf{N}$ and with $b,a \in G$
$\Rightarrow \int f_{n}(x)\,dx - \epsilon_{n}(b-a) \le t \le T \le \int f_{n}(x)\,dx + \epsilon_{n}(b-a) \ \forall n \in \mathbf{N}$
and hence
$0\le T-t \le 2\epsilon_{n}(b-a) $
now local uniform convergence $\Leftrightarrow \lim\limits_{n\to \infty}\ \sup\limits_{x\in G}\ |f_{n}-f| = 0$.
it follows that :
for $n\to \infty : T=t$
So $f$ is Riemann integrable and thus also continuous on $G$.
Proof (new version)
We show that $f$ is continuous in $G$.
Let $t_{0} \in G$, then by premise there is a environment $U$ in $t_{0}$ so that $f_{n}$ restricted to $G\cap U$ converges uniformly to $f$ restricted to $G\cap U$. Then all $f_{n}$ restricted to $G\cap U$ in $t_{0}$ are continuous (because continuity is a local property). Therefore $f$ restricted to $G \cap U$ is continuous in $t_{0}$. Again because of the local property of continuity it follows that $f$ is therefore also continuous in $t_{0}$.
$f$ is integrable on $G$ because it is continuous $G$.
Is this proof (new version) fine? Help is greatly appreciated.