I'm trying to solve the next problem: Let $f$ be Riemann integrable on the interval $I$ and $a,x\in I$. Prove that $F\left(x\right)=\int_{a}^{x}f\left(t\right)dt$ is a continuous function on $I$.
So given $c\in I$ and $\varepsilon>0$ I want to find $\delta>0$ such that if $x\in\left(c-\delta,c+\delta\right)\cap I$ then $$ \mid F\left(x\right)-F\left(c\right)\mid=\mid\int_{a}^{x}f\left(t\right)dt-\int_{a}^{c}f\left(t\right)dt\mid=\mid\int_{c}^{x}f\left(t\right)dt\mid<\varepsilon $$
The case when the function is bounded by a constant $M>0$ is easy because I can take for example $\delta=\frac{\varepsilon}{M}$. But I don't know if the function is bounded on $I$. Also if the interval is compact then Riemann integrability of $f$ implies that the function is bounded, but the interval is arbitrary and it could be open and I have problems for choose the appropriate $\delta>0$. I'm thinking that may be the problem could be wrong but I don't know. Could you help me please?
Thanks.
Yes, $f$ is bounded. That is part of the definition of Riemann-integrable function: if $f$ is Riemann-integrable on $[a,b]$ then $f|_{[a,b]}$ is bounded. And if you want to prove that $F$ is continuous at $x$, then you can assume without loss of generality that you are working on an interval $[a,b]$ with $b>x$. Unless $x$ the largest element of $I$. in which case you work on $[a,x]$.