Q: Uniform convergence of a sequence of functions given by integrals If $f_n:(0,\infty)\to \mathbb{R}$ is a sequence of functions converging uniformly to a function $f:(0,\infty)\to\mathbb{R}$, and also if each $f_n$ is Riemann-integrable over $[a,b]$ for any $0<a<b<\infty$, I know that 
$$
\lim_{n\to \infty}\int_a^bf_n(x)dx = \int_a^bf(x)dx.
$$
Fixing a $a = a_0$, we can say that the sequence of functions $F_n(x) = \int_{a_0}^xf_n(t)dt$ converges to $F(x) = \int_{a_0}^xf(t)dt$. But is this converge uniform? 
I'm tried to show that this sequence is a Cauchy sequence: given $\epsilon>0$ there exists $n_0$ such that $m,n>n_0$ implies $|\int_{a_0}^r f_m(x)dx - \int_{a_0}^rf_n(x)dx|<\epsilon$ for all $r>0$. But I don't know how to use the uniform convergence of $(f_n)$ without losing the independence on the point of the domain. Any help will be appreciated.   

A: I don't believe the statement you're trying to prove is correct; the result does not hold on the unbounded interval $(0, \infty)$. If this interval was bounded, this result would hold, as you have already noted.
For a concrete example, consider $f(x) = x + 1$ and the sequence of functions $\{f_n\}$ defined by $$f_n(x) = x + \frac{n - 1}{n}.$$ Then clearly, $\{f_n\}$ converges uniformly to $f$. On the other hand, we have $$\lvert F(x) - F_n(x) \rvert = \bigg\lvert \int_0^x f(t) \, dt - \int_0^x f_n(t) \, dt \bigg\rvert = \bigg\lvert \int_0^x \frac 1n \, dt \bigg\rvert = \frac xn$$ for all $n$, which is larger than any $\varepsilon > 0$ for sufficiently large $x$. 
