Proof of $\lim_{x\to b}\int_a^x| f(t)|dt$ exists $\implies$ $ \lim_{x\to b}\int_a^x f(t)dt$ exists. Let $f:[a,b[\to \mathbb R$ continuous and integrable on $[a,b[$. I want to prove that $\displaystyle\lim_{x\to b}\int_a^x |f(t)|dt$ exists $\implies$ $\displaystyle\lim_{x\to b}\int_a^x f(t)dt$ exists. I set $F(x)=\int_a^x f(t)dt.$ In my course they proved that $F$ is uniformly continuous on $[a,b[$ and thus $\lim_{x\to b}F(x)$ exist.

Can I do as follow : I set $f^+(x)=\max\{f(x),0\}$ and $f^{-}(x)=-\min\{f(x),0\}$. Since $f^{+}$ and $f^-$ are continuous  $$0\leq  f^+\leq |f|\quad \text{and}\quad 0\leq f^-\leq |f|,$$
and $\int_a^b |f|$ exists, we have that $$\lim_{x\to b}\int_a^x f^+\quad \text{and}\quad \lim_{x\to b}\int_a^x f^-$$
exists. Therefore, $$\lim_{x\to b}\int_a^x f=\lim_{x\to b}\int_a^x f^+-\lim_{x\to b}\int_a^x f^-,$$
and thus it exists. 

Does it work ?
It is enough to show that for every Cauchy sequence $x_n$, $\int_{a}^{x_n}f(t)dt$ is a Cauchy sequence. This results from the fact that $|\int_{x_n}^{x_m}f(t)dt|\leq \int_{x_n}^{x_m}|f(t)|dt$.
It is enough to show that for every Cauchy sequence $x_n$, $\int_{a}^{x_n}f(t)dt$ is a Cauchy sequence. This results from the fact that $|\int_{x_n}^{x_m}f(t)dt|\leq \int_{x_n}^{x_m}|f(t)|dt$.