A Riemann integral with a jump discontinuity at its lower limit I want to prove the following result.
Result: Let $f:[a,b]\to \mathbb{R}$. Assume that $f$ is continuous on $(a,b]$ and 
$$\lim_{x\to a^+}f(x)=L.$$ Then,$$\int_{a}^bf=\lim_{c\to a^+}\int_{c}^bf.$$
I tried finding this result on the Calculus books that I had but can't find it. So, I decided to prove this. I need the $\epsilon$-$\delta$ definition to tprove the limit.
My attempt:
I know that $f$ is Riemann integrable on $[a,b]$ and hence integrable on both $[a,c]$ and $[c,b]$. If $c\in(a, b)$ then 
$$\begin{align}
\left\lvert\int_c^bf-\int_a^b f\right\rvert&=\left\lvert\int_a^cf\right\rvert\\
&\leq\int_a^c|f|.
\end{align}
$$
I got stuck in here (maybe I can't find the right trick, its because $f(a)$ might not exist). I don't know how to proceed. For a given $\epsilon>0$, I don't know how to find $\delta>0$ such that whenever $0<c-a<\delta$, then
$$\left\lvert\int_c^bf-\int_a^b f\right\rvert<\epsilon.$$
I need some help. Thank you
 A: Let $\epsilon>0$. We want to show that there is some $c_\epsilon$ s.t. $\int_a^c |f|<\epsilon$.
Since $\lim_\limits{x\rightarrow a^{+}} f(x)=L$, there is some $\delta$ such that $|f(x)|<2L$ when $a<x<a+\delta$.
Then choose $c$ such that $c-a<\min \{\epsilon/2L, \delta\}$. Then $\int_a^c |f| < \int_a^c 2L < (c-a) 2L < \epsilon$.
A: Let $$F(x) = \int_{x} ^{b} f(t) \, dt$$ then by Fundamental theorem of calculus $F(x) $ is continuous on $[a, b] $ and differenrentiable on points of continuity of $f$. The continuity of $F$ at $a$ gives you the result you are seeking namely $$\int_{a} ^{b} f(t) \, dt=F(a) = \lim_{c\to a^{+}} F(c) =\lim_{c\to a^{+}} \int_{c} ^{b} f(t) \, dt$$ Note that the above result holds because of continuity of $F$ at $a$ and for this it is only required that $f$ be Riemann intgrable on $[a, b] $. We don't need to know if limit of $f$ at $a$ exists or not nor do we need the continuity of $f$ on $(a, b] $. The conditions given for $f$ in the question are sufficient to ensure that $f$ is Riemann integrable on $[a, b] $. 
