How to prove Riemann sum wrt. any point will give same result (left, right, middle, etc.)

Question

If $f(x)$ is a differentiable function. If I want to find the Riemann sum between points $a$ and $b$.
I'll divide the part of function from $a$ to $b$ into $n$ pieces. So the width of each piece will be $\Delta x=\frac{b-a}{n}$. Let the points that demarcate each part are $x_0,x_1,...,x_n$ such that $x_0=a$ and $x_n=b$, and $x_i=x_{i-1} + \Delta x$. The area of each (rectangular) piece will be $\Delta x f(x_i^*)$, such that $x_i^*$ is a point between $x_i$ and $x_{i+1}$.

I want to prove that $x_i^*$ can be any point between $x_i$ and $x_{i+1}$, and the Riemann sum (i.e. $\sum_{i=0}^n f(x_i^*)\Delta x$) as $n \to \infty$ will give the same result.

Will this kind of proof work?
$x_i\le x_i^*\le x_{i+1}$
$\Rightarrow x_i\le x_i^*\le x_i+\Delta x$
as $n\to \infty$, $\Delta x\to 0$
$\Rightarrow \lim_{\Delta x\to 0} x_i\le \lim_{\Delta x\to 0} x_i^*\le \lim_{\Delta x\to 0} x_i+\Delta x$
$\therefore x_i \le x_i^*\le x_i$
Hence $x_i^*$ can be any point between $x_i$ and $x_{i+1}$

Answer 1

Integrals are defined as the limit of the Riemann sum for any partition with distinguished points, if such a limit exists. So, the question you are asking is actually defined as being true in order for an integral to exist. You can imagine that, if the limit depended on the points $x_i^*$ that you chose, you'd get a different limit for different points and the integral wouldn't be unique.

Perhaps, though, a better way to think about this, with respect to the question you are asking is by looking at \begin{align*} s&:=\sum_{i=1}^n\inf_{x\in[x_i+\delta x]}f(x)\Delta x\\ S&:=\sum_{i=1}^n\sup_{x\in[x_i+\delta x]}f(x)\Delta x \end{align*} $\int_a^bf(x)dx$ exists only when $\lim s=\lim S$. Since $\inf_{x\in[x_i+\Delta x]f(x)}\leq f(x_i^*)\leq \sup_{x\in[x_i+\Delta x]}f(x)$, it follows that it will work for any values of $x_i^*$ that you choose.

btw, if you're unfamiliar with the notation, you can think of $\inf$ as minimum and $\sup$ as maximum, though it's a little more technial than that.

Answer 2

The riemann sum is defined as

$$\lim_{n\rightarrow \infty}\sum_{i=1}^n f(t_i+h_i)(t_{i+1}-t_i),$$

where $t_i$ is a dense partition of $[a,b]$, and $t_i+h_i\in [t_{i},t_{i+1}]$.

We use the fact that any function that is continuous is also uniformly continuous in an interval classically. This means for any $t$ and any $\epsilon$ there is $\delta$ such that for all $x$

$$|f(x)-f(t)|<\epsilon, |x-t|<\delta,$$

and the delta works for all $t$. Hence we reduce the intervals until each is within $\delta$ of each other, which happens since the partition is dense. Then

$$\lim_{n\rightarrow \infty}\sum_{i=1}^n f(t_i+h_i)(t_{i+1}-t_i),$$

$$<\lim_{n\rightarrow \infty}\sum_{i=1}^n (f(x_i)+\epsilon) (t_{i+1}-t_i)=\lim_{n\rightarrow \infty}\sum_{i=1}^n f(x_i) (t_{i+1}-t_i)+(b-a)\epsilon,$$

$$>\lim_{n\rightarrow \infty}\sum_{i=1}^n (f(x_i)-\epsilon) (t_{i+1}-t_i)=\lim_{n\rightarrow \infty}\sum_{i=1}^n f(x_i) (t_{i+1}-t_i)-(b-a)\epsilon$$

As $\epsilon$ was arbitrarily chosen, this should suffice. I think uniform continuity should be enough.