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.

  • $\begingroup$ @JackD'Aurizio yes but how does this partition stands with the limit when $t\to 1^-$? $\endgroup$ – Mickey Jul 13 '17 at 15:10
  • $\begingroup$ To the person who voted to close it for "off topic - missing context", how does this missing context or off topic? $\endgroup$ – Mickey Jul 13 '17 at 15:13
  • $\begingroup$ @JackD'Aurizio Okay, maybe I didn't explain myself properly. Derive a good partition meaning I couldn't tie the partition to the sum with the limit of $t$ going to $1$. For the "lacks any attempt on your side$, how do you know I haven't tried solving it? Again (as you have answered to another question of mine, and as I commented on your answer) - trivial is a subjective concept, what is trivial for you is not trivial for other people. $\endgroup$ – Mickey Jul 13 '17 at 15:21
  • $\begingroup$ @JackD'Aurizio Shouldn't $N\to \infty$ be taken before $t\to 1^-$? If so it is correct saying that after taking the limit on $N$, we get this expression: $\lim_{t\to 1^-}\int_0^tf$, and then taking the limit of $1$ gives as $\int_0^1f$? Does taking the limits one after the other in that fashion is correct? $\endgroup$ – Mickey Jul 13 '17 at 15:43
  • $\begingroup$ Why do people want to close this topic? MSE is weird sometimes. $\endgroup$ – zhw. Jul 14 '17 at 5:23

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.

  • $\begingroup$ Thanks for the detailed and thorough answer! $\endgroup$ – Mickey Jul 14 '17 at 21:01

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