If the function has an integral then the following limit exists? I just solved the following problem:  $f \in R[0,1]$ and the following limit exists: $\lim_{t \to 1^-}(\lim_{N \to \infty}\sum_{n=1}^{N}f(t^n)(t^n-t^{n+1}))=A$  The problem was to show that $\int_{0}^{1}f = A$.  I solved it, but since I used the fact that the limit exists in my solution, I am now wondering what could be an example for such $f$ such that the limit above does not exists? I suppose there is one, otherwise the given information that the limit exists is not needed for the problem.
 A: The existence of the limit is not needed:
First note that $\left| f \right| \in R[0,1]$. Starting at a sufficiently big $t$ the series 
$$\sum_{n=0}^\infty \left| f(t^n) \right| (t^n-t^{n+1})$$
and therefore 
$$\sum_{n=0}^\infty f(t^n) (t^n-t^{n+1})$$
exist. If they would not, $\left| f \right|$ could not be Riemann-integrable.
Now we take an arbitrary sequence $(t_n) \to 1^-$. Our goal is to try to transfer the given limit into a sequence of Riemann-sums. If we cut the series at an arbitrary $N \in \mathbb{N}$, what remains is a Riemann-sum with biggest interval size $t-t^2$
$$\sum_{n=0}^N f(t^n) (t^n-t^{n+1}).$$
So if $(t_n)$ approaches $1$, the Riemann sum gets finer and finer.
Now to be precise: For an arbitrary $n \in \mathbb{N}$ we want to define the Riemann sum $R_l$ the following way: Take $k_l$ so large that $t_{k_l} - t_{k_l}^2 < \frac{1}{l}$ and take $N$ so large that 
$$\sum_{n=N+1}^\infty f(t_{k_l}^n) (t_{k_l}^n-t_{k_l}^{n+1}) < \frac{1}{l}$$
and that we can find an $x_l \in [0,t^{N+1}]$ such that $\left| f(x_l)t^{N+1} \right| < \frac{1}{l}$ (existence of such an $x_l$ follows again from Riemann-integrability).
Now 
$$R_l = \sum_{n=0}^N f(t_{k_l}^n) (t_{k_l}^n-t_{k_l}^{n+1}) + f(x_l)t^{N+1}$$
is really a Riemann sum for the integral which is $\frac{1}{l}$-precise and
$$\left| \sum_{n=0}^\infty f(t_{k_l}^n) (t_{k_l}^n-t_{k_l}^{n+1}) - R_l \right| < \frac{2}{l}$$
As we know that $\lim_{l \to \infty} R_l = A$ we also get that $\lim_{l \to \infty} \sum_{n=0}^\infty f(t_{k_l}^n) (t_{k_l}^n-t_{k_l}^{n+1}) = A$.
As $(t_n)$ was arbitrary it follows that
$$\lim_{t \to 1^-}\sum_{n=0}^\infty f(t^n) (t^n-t^{n+1}) = A.$$
