# $\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}f(\frac{k+\theta}{n})=\int_{0}^{1}f(x)dx$ where $\theta\in(0,1)$?

Let $$f\colon [0,1]\to\mathbb{R}$$ be a continuous (or Riemann-integrable) function. As we already know, the next equation holds: $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac{k}{n}\right)=\int_{0}^{1}f(x)dx$$ Then, I want to prove the following: $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac{k+\theta}{n}\right)=\int_{0}^{1}f(x)dx\tag{1}$$ where $$\theta\in(0,1)$$. I don't know whether (1) works, but I think this equation is correct.

Now, I could prove (1) when $$f$$ is monotonically by taking the difference between two sides and dividing interval of integration at $$x=\frac{k+\theta}{n}$$. But I can't do when $$f$$ is a general function.

Can anyone teach me the proof of (1) in the general case? Thank you.

• If the integral exists, then any $\theta$ as advertised will do, simply by definition of the Riemann integral. Commented Jan 27, 2022 at 15:44

Simply from definition. You took a partition $$P\colon 0 < 1/n < 2/n < \dots < n/n = 1$$ of the interval $$[0,1]$$ where $$k/n =: x_n$$ are the partioning points, and the "signal point" are $$\xi_{k+1} := (k+\theta)/n \in [k/n, (k+1)/n] = I_{k+1}$$ where $$I_{k+1}$$ is the $$(k+1)$$-th subinterval of this partition. Then the LHS is exactly a limit of a sequence of Riemannian sums: $$\lim_{\Vert P \Vert \to 0} \sum_1^{n} f(\xi_k) \vert I_{k} \vert,$$ where $$\Vert P \Vert = \max_k (\vert I_k \vert) = 1/n \to 0$$ as $$n \to +\infty$$. Since $$f$$ is integrable, each Riemannian sum must converge to the Riemannian integral. Done.