Integrablity and continuity 
Let $g:R->R$. Assume $\int_{-\infty}^{x}g\left(u\right)du$ converges
for every $x∈R$.
Prove $G(x)=\int_{-\infty}^{x}g\left(u\right)du$ is continuous for
every $x∈R$.

I've seen a similar post in MSE with general $a,b$ instead of $-∞$ but wasn't able to the proof to my case..
I understand why it's true, but I'm not able to formalize it,
I've tried using the delta epsilon definition of continuous functions with no luck..
Is there something obvious I'm missing?

Let $x_{0}\in\mathbb{R}$. The improper integral converges for every value, thus for every $x\in\mathbb{R}$ we get:
$$\int_{-\infty}^{x}f\left(t\right)dt=\int_{-\infty}^{x_{0}}f\left(t\right)dt+\int_{x_{0}}^{x}f\left(t\right)dt\;\;\;\left(*\right)$$
Assuming, without loss of generality that $x_{0}<x$ and consider the definite integral $\int_{x_{0}}^{x}f\left(t\right)dt$.
as seen above, it is a subtraction of two converting integrals, thus $\int_{x_{0}}^{x}f\left(t\right)dt$ exists and $f$ is integrable on $\left[x_{0},x\right]$ for every $x\in\mathbb{R}$, and therefore $f$ is bounded, namely there exists $M>0$ such that for every $t\in [x_0,x]$ we have:
$$\left|f\left(t\right)\right|\le M$$
We know that $f$ is integrable $\Rightarrow$ $|f|$ is integrable, thus by the monocity of the definite integral, we have:
$$\int_{x_{0}}^{x}\left|f\left(t\right)\right|dt\le\int_{x_{0}}^{x}Mdt=M\left(x-x_{0}\right)\;\;\;\left(**\right)$$
We also know that:
$$\left|\int_{x_{0}}^{x}f\left(t\right)dt\right|\le \int_{x_{0}}^{x}\left|f\left(t\right)\right|dt\;\;\;\left(***\right)$$
Now we check the one-sided limit on the right. Let $\varepsilon>0$.
We choose $\delta=\frac{\varepsilon}{M}$ and let $x\in\mathbb{R}$ such that $x-x_{0}<\delta$. then:
$$\left|F\left(x\right)-F\left(x_{0}\right)\right|=\left|\int_{-\infty}^{x}f\left(t\right)dt-\int_{-\infty}^{x_{0}}f\left(t\right)dt\right|\underset{\left(*\right)}{=}\left|\int_{x_{0}}^{x}f\left(t\right)dt\right|\underset{\left(***\right)}{\le}\int_{x_{0}}^{x}\left|f\left(t\right)\right|dt\underset{\left(**\right)}{\le}M\left(x-x_{0}\right)<\varepsilon$$
You can easily check for $x<x_0$
