If $\int_{a+}^bfdx$ exists, then show that $\int_{a+}^cfdx$ exists for any $c\in(a,b)$ and $\lim_{c\to a+}\int_{a+}^cfdx=0$.
I proved this as follows.
(i)By Cauchy's criterion, for given $\varepsilon>0$, there is $\delta>0$ such that $a<r\le s<a+\delta$ implies that $|\int_r^sfdx|<\varepsilon$. Fix $c\in(a,b)$. We have $$ |\int_r^cfdx-\int_s^cfdx|=|\int_r^sfdx|<\varepsilon. $$ It follows from Cauchy's criterion again that $\int_{a+}^cfdx$ exists for any $c\in(a,b)$.
(ii)Let $I=\int_{a+}^cfdx$. For given $\varepsilon>0$, there is $\delta>0$ such that if $a<d<a+\delta$ then $|I-\int_d^cfdx|<\frac{\varepsilon}{2}$. Fix $d\in(a,a+\delta)$. Since $f$ is Riemann integrable on $[d,c]$, $f$ is bounded on $[d,c]$, i.e. there is $M>0$ such that $|f(x)|\le M$ for all $x\in[d,c]$.
Now, letting $\delta'=\min(\delta,\frac{\varepsilon}{2M})$ and taking $d'<d$ with $c<d'<a+\delta$, we see that if $a<c<a+\delta'$ then $$ |I|\le|I-\int_{d'}^cfdx|+|\int_{d'}^cfdx|<\frac{\varepsilon}{2}+M\delta'<\varepsilon. $$
Is this argument correct? Any help will be greatly appreciated.