Riemann definition of integrability to show integral exists

Question

$\textbf{Problem Statement :}$ Let $f$ be a real valued function defined on $\mathbb{R}$. Suppose that $\lim_{n \rightarrow \infty} \sum_{i = 1}^n \frac{1}{n}f\left(\frac{i}{n}\right)$ exists and is equal to two.

The first part of this question asks whether $\int_0^1 f(x)$ exists and if it exists, whether it must be equal to $2$, and the second part of this question asks you to consider the same question when $f$ is continuous.

In the first part, I think it is not necessary that the integral exists nor must it be equal to two if it does. We can say that the function is unbounded on the irrationals as a case to when the integral does not exist, and take a function which is 0 on all irrationals but whatever it needs to be on the rationals so the limit of that sum converges, as this integral would be equal to $0$.

However when the function is continuous, we are certain that this integral must exist, since by continuity, $\lvert U_n(f,P) - L_n(f,P)\rvert \leq \lvert \sum_{i=1}^n (M_i - m_i)\frac{1}{n}\rvert \leq \epsilon \sum_{i = 1}^n \frac{1}{n} = \epsilon$. To show that the value is $2$, we see that $m_i \leq f\left(\frac{1}{n}\right) \leq M_i$. However here, I have some trouble since I cannot use a 'reverse squeeze theorem' to claim that $\lim_{n \rightarrow \infty} \vert U_n(f,P) - 2 \vert < \epsilon$ or same with lower partition. If someone could give me a slight nudge on how to proceed, I would be grateful.

Answer 1

For $f=2\chi_{{\bf{Q}}}$ on $[0,1]$, $f$ is not Riemann integrable but $\displaystyle\sum_{i=1}^{n}\dfrac{1}{n}f\left(\dfrac{i}{n}\right)=\sum_{i=1}^{n}\dfrac{2}{n}=2\rightarrow 2$.

Answer 2

You don't need unbounded functions because by default Riemann integration assumes bounded functions. Further the limit in question is a limit of Riemann sum. And then if $f$ is integrable its integral must be equal to the given limit $2$.

But the existence of given limit can't guarantee the integrability of the function involved. Why?? The Riemann integrability of the function involves two things: arbitrary partitions of the interval of integration such that norm of the partition tends to $0$ and function values at arbitrary tag points to make the Riemann sum. Your limit assumes a specific partition (subintervals of equal length) and also a specific choice of tag points (right end points of subintervals created by the partition). In general one of these things must be arbitrary if one wants to ensure integrability. A counter-example in another answer shows why fixing both the choice of partitions and tag points won't work.