# Show that Riemann sum does not depend on rectangle endpoint choice

To calculate a Riemann sum by assuming right endpoints, we have $$S= \lim_{n\to\infty} { \sum_{i=1}^{n} f(x_i^*)\Delta x } = \lim_{n\to\infty} { \sum_{i=1}^{n} f\left( i\Delta x+ a\right)\Delta x .}$$

Then, to calculate $$S$$ using any endpoint, I introduce a parameter $$\alpha, \quad 0\leq\alpha\leq1$$ such that $$S = \lim_{n\to\infty} { \sum_{i=1}^{n} f\left( (i-\alpha)\Delta x+ a\right)\Delta x }.$$ For $$\alpha = 0$$, we have right endpoints, for $$\alpha = 1$$ we use left endpoints, for $$\alpha = \tfrac{1}{2}$$ we use mid point, and so on.

How could I prove that $$S$$ does not depend on $$\alpha$$ for any integrable $$f$$ analytically? I have done so numerically, but am struggling with the analytical method. Since I am not working with a particular function, I can't use the classic textbook formulas for the series for $$\{i^n\}$$. I would be inclined to try to show that partial derivative wrt $$\alpha$$ is 0, but I'm not sure under which conditions I can pass a derivative operator through an infinite sum.

Are there any helpful documents/references/answers you could give me?

EDIT: mistake.

## 1 Answer

A good reference is Wikipedia Riemann integral article.

The article mentions two definitions of Riemann integral and proves that those are equivalent.

What you ask is related to that. The proof goes through Darboux integral.

It only supposes that the function $$f$$ is Riemann integrable. Easier proves can be done with further assumptions of $$f$$ like you devise, i.e. if $$f$$ is supposed to be continuously differentiable.