Suppose we have the original Riemann sum with no removed partitions, where $f(x)$ is continuous and reimmen integratable on the closed interval $[a,b]$. $$\lim_{n\to\infty}\sum_{i=1}^{n}f\left(a+\left(\frac{b-a}{n}\right)i\right)\left(\frac{b-a}{n}\right)$$

If we remove $s$ partitions for every $d$ partitions in the interval $[a,b]$ and add the remaining partitions as $n\to\infty$ the resulting sum is

$$\lim_{n\to\infty}\sum_{i=1}^{{\left(d-s\right)}\lfloor\frac{n}{d}\rfloor+\left(n\text{mod}{d}\right)}f\left(a+\left(\frac{b-a}{n}\right)s(i-g_1)+g_2\right)\left(\frac{b-a}{n}\right)$$

Where $S(i)$ is a piece-wise linear vector that skips $s$ for every $d$ partitions. For example if we skip one partition out of every four partitions ,instead of the vector $i$ whose outputs are ($1$,$2$,$3$,$4$,$5$...), we have $s(1)=1$, $s(2)=3$, $s(3)=4$, $s(4)=5$, $s(5)=7$,$s(6)=8$...).

So for in my theorem I'm trying to show that

$$\lim_{n\to\infty}\sum_{i=1}^{{\left(d-s\right)}\lfloor\frac{n}{d}\rfloor+\left(n\text{mod}{d}\right)}f\left(a+\left(\frac{b-a}{n}\right)s(i-g_1)+g_2\right)\left(\frac{b-a}{n}\right)=$$ $$\frac{d-s}{d}\lim_{n\to\infty}\sum_{i=1}^{n}f\left(a+\left(\frac{b-a}{n}\right)i\right)\left(\frac{b-a}{n}\right)=\frac{d-s}{d}\int_{a}^{b}f(x)$$

I know as all the partitions of the orginal sum ($\lim_{n\to\infty}\sum_{i=1}^{n}f\left(a+\left(\frac{b-a}{n}\right)i\right)\left(\frac{b-a}{n}\right)$) come closer to being equal, the sum of the fraction of remaining partitions will be the same as that fraction of the orginal reimmen sum.

To prove the partitions of original reimmen sum comes closer to being equal I found the following.

$$\lim_{n\to\infty}f\left(a+\left(\frac{b-a}{n}\right)\right)\left(\frac{b-a}{n}\right)<\frac{\lim_{n\to\infty}\sum_{i=1}^{n}f\left(a+\left(\frac{b-a}{n}\right)i\right)\left(\frac{b-a}{n}\right)}{n}<\lim_{n\to\infty}f(b)\left(\frac{b-a}{n}\right)$$

And

$$\lim_{n\to\infty}f(b)\left(\frac{b-a}{n}\right)-\lim_{n\to\infty}f\left(a+\left(\frac{b-a}{n}\right)\right)\left(\frac{b-a}{n}\right)=0$$

Am I on the right direction with proving this? If not can you give expand on a better way of proving this?

**EDIT:**
**I did posted my incomplete answer but its cluttered. Is there a simpler (and more rigorous proof) that can be done?**

SECOND EDIT:

The person who answered my question deleted his post for unknown reasons. He has sent no reply as to why he did so. I posted my version of his answer down below my incomplete answer. I am waiting for another answer that expands or gives a better proof.

Third edit: I deleted my original proof. Christian Blatters answer remains and there is a new answer from another user but Im not sure if its correct.