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.
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.