Definition of improper Riemann integral My lecture notes have defined the improper Riemann integral as follows

Let $f : (a,b] \rightarrow \mathbb{R}$ be a function such that $f|_{[\tilde a,b]}$ is Riemann integrable on every closed interval $[\tilde a, b]$ with $a < \tilde a < b.$ $f$ is improperly Riemann integrable if, for any sequence $(a_n)_n \subseteq (a,b]$ that converges to $a$, the sequence $\Big( \int_{a_n}^b f(x)\, \mathrm{d}x \Big)_{n \in \mathbb{N}}$ converges. In that case, the (common) limit is written $\int_a^b f(x) \, \mathrm{d}x.$

Is it enough if the sequence of $\int_{a_n}^b f(x) \, \mathrm{d}x$ converges for any single fixed sequence $(a_n)_n \rightarrow a$? I think it should be, because $$F(y) := \int_y^b f(x) \, \mathrm{d}x$$ defines a continuous function on $(a,b]$, right?
 A: No. Let's say $(a,b]=(0,1]$ and $f$ on each of the subintervals $[\frac1{n},\frac1{n-1}]$ has two spikes that connect on the $x$-axis in the middle of the subinterval: positive one looking like this /\ of height $n^3$, and negative one is its reflection about the $x$-axis. If you take $a_n=\frac1{n}$ then $\int_{\frac1{n}}^1f(x)dx=0$ for all $n$, because on each $[\frac1{k},\frac1{k-1}]$ for $k\leq n$ integrals over the two spikes cancel each other. However, if you cut off in the middle of the $n$-th subinterval then for $k<n$ the integrals over the subintervals still cancel, but not over the half of the $n$-th one. So $\int_{\frac1{n-1/2}}^1f(x)dx$ is the area under the positive $n$-th spike, i.e. $\frac12(\frac1{n-1}-\frac1{n})\,n^3=\frac{n^3}{2n(n-1)}\to\infty$ when $n\to\infty$. Therefore, the improper integral $\int_{0}^1f(x)dx=0$ doesn't exist.
The function $F(y)$ is continuous on $(a,b]$, but not necessarily at $a$. In this case $F(\frac1{n})$ converges to $0$, but $F(\frac1{n-1/2})$ goes to infinity, so $F$ isn't continuious at $0$.
Convergence on a single sequence will be enough if the function $f$ is positive because then $F(y)$ is monotone. For any other sequence $\widetilde{a}_k\to a$ you can find $a_{n_k}$ so that $a_{n_k}\to a$ and $a<\widetilde{a}_k\leq a_{n_k}$. Then $\lim_n F(a_n)\geq F(\widetilde{a}_k)\geq F(a_{n_k})$ and $F(\widetilde{a}_k)$ converges to the same limit by the squeeze theorem.
