$\textbf{I would like a proof in detail of the following question.}$ 

Let $(x_n)$ be an sequence in $[a,b)$ converging to $b$. We will show that $f(x_n)$ is a converging sequence, and since the sequence $x_n$ was arbitrary, this means the limit of $f$ at $b$ exists.

Let $y_n:= \int_a^{x_n}f(x)dx$. We will show this is a Cauchy sequence of numbers in $\mathbb{R}$, and hence, convergent.

Let $M$ be the bound of $f$ in $[a,b)$.

$|y_n-y_m| =|\int_{x_m}^{x_n}f(x)dx| \leq \int_{x_m}^{x_n}|f(x)|dx \leq M|x_n-x_m|$

Since $(x_n)$ is a convergent sequence, it is also Cauchy, meaning the differences $|x_n-x_m|$ can be made arbitrarily small provided $n,m$ are large enough. This shows $(y_n)$ is also Cauchy, so it converges and the limit of $f$ at $b$ exists.

Then, we can define $F(b) = \int_{a}^{b}f(x)dx := \lim_{n\rightarrow\infty} \int_{a}^{x_n}f(x)dx$