# 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

• Observe that since limit of $f$ exists, when $x$ is near $a$, $f(x)$ is roughly $L$, by the definition of limit. – k99731 Dec 28 '16 at 5:55
• @k99731 Yeah I know that one. But how can we proceed then? I will be glad if you provide an answer.:D – Juniven Dec 28 '16 at 5:59

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 $[a, b]$.
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 $\int_a^c |f| < \int_a^c 2L < (c-a) 2L < \epsilon$.
• Unless we have $L>0$, then it makes sense – Juniven Dec 28 '16 at 6:13
• @juniven Well you can just add absolute sign to $L$. – k99731 Dec 28 '16 at 6:18
• What if $L=0$, how can you repair your proof? – Juniven Dec 28 '16 at 6:23
• @juniven In that case, $|f(x)|<1$ for some $\delta$. Then $\int_a^c |f(x)|<c-a$. We just have to choose $c$ such that $c-a<\min\{\epsilon, \delta\}$. – k99731 Dec 28 '16 at 6:28