Uniform convergence and improper Riemann integral of the second kind It is well  known that the property:

If $f_n:A\to \mathbb{R}$ is Riemann integrable for all $n$ and $(f_n)$ converges uniformly to $f:A\to\mathbb{R}$, then $f$ is Riemann integrable.

is false in general  when $A$ is of the form $[a,\infty)$. For example:
$$
f_n(x) = \left\{
    \begin{array}{ll}
        1/x & \mbox{if } 1\le x\le n\\\
        n/x^2 & \mbox{if }  x\ge  n
    \end{array}
\right.
$$
What about the case $A=(a,b)$?  Can we still  construct a counter-example?
 A: No, we cannot construct a counterexample. First of all, we know that $f$ is integrable over any compact $[c,d] \subset (a,b)$, since this result is true for this case - as already mentioned in the comments. Now fix $n \in \mathbb{N}$ such that $\|f_n -f\|_{(a,b)} < \epsilon$, then for any $0<h_1<h_2< \min(a-b,1)$ we have
$$\tag{1}\Big|\int_{a+h_1}^{a+h_2} f \, \mathrm{d} x -\int_{a+h_1}^{a+h_2} f_n \, \mathrm{d} x \Big| \leq (h_2-h_1) \|f_n -f\|_{(a,b)} < \epsilon.$$
Because $\mathbb{R}$ is complete, we know that being convergent and Cauchy is the same. So, since $f_n$ is improper integrable, we know that there exists a $\delta >0$ (w.l.o.g. $\delta < \min(1,b-a)$) such that for any $0<h_1<h_2 < \delta$ we have
$$\Big| \int_{a+h_1}^{a+h_2} f_n \, \mathrm{d} x \Big| < \epsilon.$$
Both together gives
$$ \Big| \int_{a+h_1}^{a+h_2} f \, \mathrm{d}x \Big| < 2 \epsilon$$
for all $0< h_1 < h_2 < \delta$. So $f$ is improver integrable in the endpoint $a$. The same argument shows the integrability for the other endpoint. Therefore, $f$ is also improper integrable.
Note that we have used in (1) that the interval $(a,b)$ is bounded. This arguemnt doesn't work for unbounded intervals.
