proving $\int_{0}^{1}f=A$ when $f:\left [ 0,1 \right ] \rightarrow \mathbb{R}$ Given $f:\left [ 0,1 \right ] \rightarrow  \mathbb{R}$ integrable function, and $\lim_{t\rightarrow 1^-}\lim_{N\rightarrow \infty }\sum_{n=1}^{N}f(t^n)(t^n-t^{n+1})=A$,
exists and is finite.
Need to prove: $\int_{0}^{1}f=A$.
Any help?   
 A: Note the assumptions carefully. In addition to assuming that $f$ is (presumably) Riemann integrable, it is assumed that 
$s_t = \lim_{N\rightarrow \infty }\sum_{n=1}^{N}f(t^n)(t^n-t^{n+1})$ exists for all $t$ sufficiently close to one, and that
$\lim_{t\uparrow 1} s_t=A$.
We want to show that $A= \int_0^1 f$.
Since $f$ is Riemann integrable, it is bounded, say $|f(t)| \le M$.
Let $t \in (0,1)$ and $P_{N,t}$ be the partition $(0,t^{N+1},t^N,t^{N-1},...,t^2,t,1)$. It is easy to see that
$\operatorname{mesh} P_{n,t} = \max(1-t,t^{N+1})$.
Let $\epsilon>0$, then since $f$ is Riemann integrable there exists some $\delta$ such that if $\operatorname{mesh} P< \delta$, then
$U(f,P)-L(f,P) < \epsilon$ (and we have $L(f,P) \le \int_0^1 f \le U(f,P)$, of course).
Choose $t',N'$ such that $\operatorname{mesh} P_{N,t} <\delta$ whenever $N \ge N', t \in (t',1)$.
Note that $L(f,P_{N,t}) \le f(t^{N+1}) t^{N+1}+\sum_{n=1}^{N}f(t^n)(t^n-t^{n+1}) +f(1) (1-t)\le U(f,P_{N,t})$, and so we have
$|f(t^{N+1}) t^{N+1}+\sum_{n=1}^{N}f(t^n)(t^n-t^{n+1}) +f(1)(1-t)- \int_0^1 f | < \epsilon$. Taking the limit as $N \to \infty$ (and remembering that $|f(t^{N+1}) t^{N+1}| \le M t^{N+1} \to 0$) gives
$|s_t+f(1)(1-t)-\int_0^1 f| \le \epsilon$.
Now taking the limit as $t \uparrow 1$ gives $|A-\int_0^1 f| \le \epsilon$. Since $\epsilon>0$ was arbitrary, we have the desired result.
A: For each $t$, the sum is a Riemann sum. It estimates $\int_{t^{N+1}}^t f(x) dx$ using the right endpoint of each rectangle. Assuming $f$ is Riemann integrable, the sum will converge to $\int_0^t f(x) dx$ as $N \to \infty$.  So it is enough to show that $\lim_{t \to 1^-} \int_0^t f(x) dx = \int_0^1 f(x) dx$. This follows from the fact that indefinite integrals are continuous, which follows from the fact that they are differentiable, which follows from the Fundamental Theorem of Calculus.
There is a technicality that I skipped here. The left endpoint of the Riemann sums is not constant in this problem, which means you technically cannot use the usual result about Riemann integration directly. But you can write:
\begin{eqnarray*}
& &\left | \sum_{n=1}^N f(t^n) \left ( t^n - t^{n+1} \right ) - \int_0^t f(x) dx \right |  \\
& \leq & \left | \sum_{n=1}^N f(t^n) \left ( t^n - t^{n+1} \right ) - \int_{t^{N+1}}^t f(x) dx \right | + \left | \int_0^{t^{N+1}} f(x) dx \right |
\end{eqnarray*}
Then the first term goes to zero by the usual result, and you can argue separately that the last term goes to zero.
A: By definition of the Riemann integral, we need a partition sequence $\Delta_k=\{0=x_{0}<x_1<x_2<\ldots <x_{n_k}=1\}$ with $\limsup_{k\to\infty}|x_{j_k}-x_{(j-1)_k}|\to 0$ and then the integral is
$$\int_0^1f=\lim_{k\to\infty}\sum_{i=1}^{n_k}f(x_{i_k})(x_{i_k}-x_{(i-1)_k})$$
Fix $0<t<1$ and let $\Delta_N=\{0=x_0, t^N, t^{N-1},\ldots ,t^2, t, t^0=1\}$. Clearly this is listed in ascending order.
Then noting that $\limsup_{n\to\infty}|t^{n}-t^{n+1}|\le t$ for $n>0$ we see that as $N\to\infty$ that
$$f(1)(1-t)+\lim_{N\to\infty}\left(\sum_{i=1}^{N}f(t^n)(t^{n}-t^{n+1})+f(t^N)(t^N-0)\right)\approx \int_0^1f$$
The last term individually goes to $0$ since $f$ is bounded and $t^N\to 0$ as $n\to\infty$. Letting $t\to 1^-$ we get the size of the mesh going to $0$ and the first term going to $0$ because $f$ is bounded, and we achieve equality.
