Let $f\in\mathfrak R[0,1],\tau_n=\{0,\frac1n,\ldots,\frac nn\}$, & $\lim_{n\to\infty}U[f;\tau_n]=\lim_{n\to\infty}L[f;\tau_n]=A$. Prove $\int_0^1f=A$

Question

Let $f\in\mathfrak R[0,1], \sigma_n = \{0,{1\over n}, \ldots, {n\over n}\}$, and $\displaystyle \lim_{n\to\infty}U[f;\sigma_n] = \lim_{n\to\infty}L[f;\sigma_n]=A$. Prove $\displaystyle\int_0^1f = A.$

First and foremost, there is already an answer here. I'm having some trouble filling in some of the details of the proof. That question I have linked to does not suppose that $f$ is Riemann integrable (i.e., $f\in\mathfrak R[0,1]$) over $[0,1]$. However, I am given that $f$ is, in fact, Riemann integrable. Does that let me skip immediately to the end of the proof and say, for all $n\in\mathbb N$, we have: $$L[f;\sigma_n]\le\sup_nL[f;\sigma_n]\le\underline{\int_0^1}f = \int_0^1f =\overline{\int_0^1}f \le \inf_nU[f;\sigma_n] \le U[f;\sigma_n],$$ which since limits preserve inequalities, and since $\displaystyle \int_0^1f$ does not depend on $n$, $$A = \lim_{n\to\infty}L[f;\sigma_n]\le\int_0^1f\le\lim_{n\to\infty}U[f;\sigma_n]=A \implies \int _0^1f=A?$$ It seems logically sound, but have I used $\sigma_n=\{0,{1\over n},\ldots,{n\over n}\}$ anywhere? Or is the condition here just to make things work "nicely"? Because I understand that $\sigma_n$ is a subdivision of $[0,1]$, and this generates a sequence for all $n\in\mathbb N$. So $\displaystyle \lim_{n\to\infty}\sigma_n = [0,1]$. But is this necessary to acknowledge in the proof anywhere?

Answer 1

I think your problem is trivial given the fact that $f$ is Riemann integrable on $[a, b]$. We have the following result in order:

Theorem: Let $f$ be a bounded function on $[a, b]$ and $P_{n}$ be a sequence of partitions of $[a, b]$ such that norm of $P_{n}$ (i.e. length of largest sub-interval created by $P_{n}$, denoted by $|P_{n}|$) tends to $0$ as $n \to\infty$. Then $$\lim_{n \to \infty}L(f;P_{n}) = \underline{\int_{a}^{b}}f(x)\,dx,\,\lim_{n \to \infty}U(f;P_{n}) = \overline{\int_{a}^{b}}f(x)\,dx$$

(for a proof see this answer)

In your case we have $P_{n} = \sigma_{n}$ and norm $|P_{n}| = 1/n$ which tends to $0$ as $n \to \infty$. It is given that $\lim_{n \to \infty}L(f;P_{n}) = \lim_{n\to \infty}U(f;P_{n}) = A$ (by the above theorem this alone guarantees that $f$ is Riemann integrable) and it is also given that $f$ is Riemann integrable (which is now redundant) so it is clear that $\int_{0}^{1}f(x)\,dx = A$.

The specific nature of points of partition $\sigma_{n}$ does not matter. All that we need is that norm of $\sigma_{n}$ tends to $0$.

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Note that the result linked in your question is different from the theorem above. It does not specifically say that norm $|P_{n}| \to 0$, but rather it says that if for some sequence of partitions $P_{n}$ we have $\lim L(f;P_{n}) = \lim U(f;P_{n})$ then $f$ is Riemann integrable. The theorem in this answer does not say anything about $f$ being Riemann integrable. The answer linked above deals with both these aspects and you should have a look at it to gain more understanding.