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?
 A: 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 $x<x_0$
