Limit of infinite Riemann sums Let $f:[0,1]\to \mathbb R$ be an integrable function.
It is also known that the next limit exists and is finite, and is
equal to $A$: $$\lim_{t\to
 1^{-}}\lim_{N\to\infty}\sum_{n=1}^{N}f(t^n)(t^n-t^{n+1})=A$$

Prove: $$\int_0^1f=A$$

$$$$
Well the sum looks similar to a Riemann sum, and the $\lim_{t\to1^-}$ is similar to Abel's summation, but for the Riemann sum part, I couldn't derive a good partition to the interval $[0,1]$ and didn't know how to use Abel's summation to solve this.
 A: Here is what is really happening in this problem: Let's extend the definition of a partition of $[0,1]$ to allow, in addition to the finite partitions that we know and love, sets $P$ of the form $P = \{a_n: n=0,1,2, \dots \},$ where $1=a_0>a_1 > a_2 > \cdots \to 0.$ Given such a $P,$ define $\mu(P)= \min \{a_n- a_{n+1}: n= 0,1,2,\dots\}.$
Suppose $f$ is bounded on $[0,1].$ If we have an "infinite partition" $P$ as above, then we can define
$$S(P,f) = \sum_{n=0}^\infty f(a_n)(a_n-a_{n+1}).$$
This series converges, in fact absolutely.
Claim: If $f$ is Riemann integrable on $[0,1],$ then
$$\lim_{\mu(P)\to 0} S(P,f) = \int_0^1 f(x)\, dx,$$
where the limit is taken over infinite partitions $P$ of $[0,1].$
None of this should be surprising. I'm going to leave the proof of this to the reader. Basically, you compare these fancier sums $S(P,f)$ to the finite Riemann sums you already know. It's pretty straightforward, but there is some bookkeeping to do.
So in the problem at hand, we would take $a_n = t^n, n=0,1,2,\dots$ Call this infinite partition $P_t.$ Then $\mu(P_t) = 1-t.$ It follows from the claim that
$$\lim_{t\to 1^-} S(P_t,f) = \int_0^1 f(x)\,dx.$$
This is the desired conclusion.
