Let $a_n$ be a strictly monotonic sequence such that $a_n \rightarrow b$. Moreover, $a_0 = a$ and $a<b$. If $f[a,b] \rightarrow \mathbb{R}$ is a bounded function, and $f$ is integrable on $[a_n, a_n+1]$ for all $n\ge0$, is $f$ integrable on $[a,b]$?
The usual examples are of combining finitely many intervals over which $f$ is Riemann integrable, however, could this be extended to the case above?
On
In each $[a_n,a_{n+1}]$, $f$ is continuous a.e. : the (Lebesgue) measure of points of discontinuity in that interval is zero. (This is Lebesgue's criterion for Riemann integrability -- $f$ bounded on $[a,b]$ is Riemann integrable on that interval if and only if it is continuous a.e. on that interval.) The union of the points of discontinuity over the intervals in your partition is a countable union of measure zero sets so has measure zero (via subadditivity of measure). So $f$ is continuous a.e. on $[a,b]$ and therefore is Riemann integrable on $[a,b]$.
You can show that $\int_a^bf=\lim_{x\to b}\int_a^x f$. In which case you can potentially extend the definition assuming that for $n$ large you can make $\int_{a_n}^xf$ as small as you wish where $x$ is arbitrary in $(a_n,a_{n+1}].$ For then you can show that $\lim_{x\to b}\int_a^xf=\lim_{k\to\infty}\int_{a}^{a_k}f=\lim_{k\to\infty}\sum_{n=0}^k\int_{a_n}^{a_{n+1}}f$.
If you wished to use Riemann sums instead of just summing the integrals then for all $n$ you could choose partition $P_n$ of $[a_n,a_{n+1}]$ such that $U(f,P_n)-\int_{a_n}^{a_{n+1}}f<\epsilon/2^{n+1}$ where $\epsilon $ is arbitrary positive.