Let $ -\infty \le a < b \le \infty $. Function $ f:(a,b) \rightarrow R $ can be integrated in the sense of Riemann on every dense $ [c,d] \subseteq (a,b) $. The integral $ \int_{a}^{b} f(x) dx $ is convergent. Show that $$ F(x) = \int_{a}^{x} f(x) dx $$ is continuous. Is it uniformly continuous?
We have $F(a) = 0$. From the fundamental theory of calculus $F(x)$ is continuous on $[a,x], x<b$. We have a left-side sonvergence in $b$ from definition, so $F$ is continuous on $[a,b]$.
Is it enough to prove continuity? How do I show uniform continuity?
This is a homework assignment.
Here are some suggestions:
Since $f$ is Riemann integrable on any interval $[c,d] \subset (a,b)$ it is bounded locally.
To prove uniform continuity of $F$, the easy case is where $f$ is bounded globally on $(a,b)$ with $|f(x)| \leqslant M$ for all $a < x < b$.
The other case is where $f$ is unbounded at one or both of the endpoints. For example, suppose $f$ is unbounded at $x = b$ and bounded and Riemann integrable on $[a,c)$ for all $c < b$.
By hypothesis, the improper integral over $(a,b)$ is convergent. By the Cauchy criterion, for any $\epsilon > 0$ there exists $C < b$ such that if $C \leqslant x < y < b$ we have
$$|F(y) - F(x)| = \left|\int_x^y f(t) \, dt \right| < \frac{\epsilon}{2}.$$
On the interval $[a,C]$, $f$ is bounded and there exists $M_C > 0$ such that $|f(x)| \leqslant M_C $ for all $x \in [a,C]$.
Now see if you can complete the proof by examining the three cases (1) $x < y < C$, (2) $x < C \leqslant y$, and (3) $C < x < y$.