Help with proving a statement based on Riemann sums?

Question

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 $s(1)=1$, $s(2)=3$, $s(3)=4$, $s(4)=5$, $s(5)=7$,$s(6)=8$...).

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

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

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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?

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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.

Answer 1

Let's simplify and assume $f$ is Riemann integrable on $[0,1].$ Fix $d,s \in \mathbb N,0<s<d.$ Consider the uniform partition of $[0,1]$ into subintervals of length $1/n,$ thinking of $n$ here as being larger than $d.$ Choose $c_i \in [(i-1)/n,i/n].$

The key to this is to look at the indices $i$ in groups of $d.$ The first group is $\{1,\dots , d\}.$ Let $A_1\subset \{1,\dots , d\}$ have $d-s$ elements. Set $I_1= [0, d/n],$ and let $M_1= \sup_{I_1}f.$ We then have the upper estimate

$$\sum_{i\in A_1} f(c_i)\frac{1}{n} \le \sum_{i\in A_1} M_1\frac{1}{n} = (d-s)M_1\frac{1}{n} = \frac{d-s}{d} M_1\frac{d}{n}.$$

We can do the same thing on the next $d$-block of indices, $\{d+1,\dots , 2d\},$ and so on. There will be $\lfloor n/d \rfloor$ $d$-blocks of indices in all. If for $k=2,\dots ,\lfloor n/d \rfloor$ we define $A_k,I_k,M_k$ in the analogous way, we get

$$\sum_{k=1}^{\lfloor n/d \rfloor} \sum_{i\in A_k} f(c_i)\frac{1}{n} \le \frac{d-s}{d}\sum_{k=1}^{\lfloor n/d \rfloor} M_k\frac{d}{n}.$$

There is the annoying last interval $I' =[\lfloor n/d \rfloor d, 1]$ to discuss. We don't know how many $f(c_i)(1/n)$ get counted here, but let's call the number $N'.$ All we can say for sure is $0\le N'\le d-s.$ Let $M'= \sup_{I'} f.$ Then, summing over these $i,$ we get

$$\sum f(c_i)\frac{1}{n} \le M'N'\frac{1}{n}.$$

Let's write the last term as

$$\frac{d-s}{s}M'(1-\lfloor n/d \rfloor d) + [M'N'\frac{1}{n} - \frac{d-s}{s}M'(1-\lfloor n/d \rfloor d)].$$

Denote the expression in brackets by $B_n.$ Verify that $B_n \to 0.$

Putting this all together gives, for the full "partial Riemann sum" we are considering,

$$\sum f(c_i)\frac{1}{n} \le \frac{d-s}{d}\left (\sum_{k=1}^{\lfloor n/d \rfloor} M_k\frac{d}{n} + M'(1-\lfloor n/d \rfloor d)\right ) + B_n.$$

Inside the parentheses we have $U(P_n,f)$ for a partition $P_n$ different from the uniform partition we started with. Since the mesh-size of $P_n$ tends to $0,$ these upper sums converge to $\int_0^1 f,$ by standard Riemann integration theory. Recalling $B_n \to 0,$ we see that the expression on the right $\to \frac{d-s}{d} \cdot \int_0^1 f.$ Therefore

$$\limsup_{n\to \infty} \sum f(c_i)\frac{1}{n} \le \frac{d-s}{d}\int_0^1 f.$$

A similar argument, using $m_k = \inf_{I_k}$ etc, gives

$$\frac{d-s}{d}\int_0^1 f \le \liminf_{n\to \infty} \sum f(c_i)\frac{1}{n}.$$

This gives the result.

Answer 2

For Riemann integrable function $f$ ,the proof hinges on three facts:

1. The theorem is valid for step function: Because removing any fractional number of partitions from a number of partitions of equal size results in the value of the integral being reduced by the same fraction as the size of the partition tends to zero.

2. limit of upper Riemann sum = limit of lower Riemann sum

3. Any Riemann intergable function $f$ is the limit of step functions {$l_m$} almost everywhere : $ \lim\limits_{m\mapsto \infty} l_m=f$ $\quad$ almost everywhere

   $ \lim\limits_{m\mapsto \infty} \int l_mdx=\int fdx$

Let {$l_m$} be a sequence step functions that converges to $f$ almost everywhere and also approximate the integral $\int_a^bfdx$ with a lower riemann sum and let {$u_m$} be a sequence of step functions that converges to $f$ almost everywhere and also approximate the integral $\int_a^bfdx$ with an upper riemann sum : That is there is some integer $m>0$ and $\epsilon>0$ such that

$\int u_mdx \ge \int fdx \ge \int l_mdx $

$|\int u_mdx-\int l_mdx| \le \epsilon$ $\quad$ based on fact 2

$|\int fdx-\int l_mdx| \le \epsilon$

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

Using fact 3 : $$\lim_{n\to\infty}\sum_{i=1}^{{\left(d-s\right)}\lfloor\frac{n}{s}\rfloor+\left(n\text{mod}{s}\right)}l_m\left(a+\left(\frac{b-a}{n}\right)s(i-g_1)+g_2\right)\left(\frac{b-a}{n}\right) \le$$ $$\lim_{n\to\infty}\sum_{i=1}^{{\left(d-s\right)}\lfloor\frac{n}{s}\rfloor+\left(n\text{mod}{s}\right)}f\left(a+\left(\frac{b-a}{n}\right)s(i-g_1)+g_2\right)\left(\frac{b-a}{n}\right) \le$$ $$\lim_{n\to\infty}\sum_{i=1}^{{\left(d-s\right)}\lfloor\frac{n}{s}\rfloor+\left(n\text{mod}{s}\right)}l_m\left(a+\left(\frac{b-a}{n}\right)s(i-g_1)+g_2\right)\left(\frac{b-a}{n}\right) + \epsilon$$

And

$$\frac{d-s}{d}\lim_{n\to\infty}\sum_{i=1}^{n}l_m\left(a+\left(\frac{b-a}{n}\right)i\right)\left(\frac{b-a}{n}\right) \le $$ $$ \frac{d-s}{d}\lim_{n\to\infty}\sum_{i=1}^{n}f\left(a+\left(\frac{b-a}{n}\right)i\right)\left(\frac{b-a}{n}\right) \le $$ $$\frac{d-s}{d}\lim_{n\to\infty}\sum_{i=1}^{n}l_m\left(a+\left(\frac{b-a}{n}\right)i\right)\left(\frac{b-a}{n}\right) + \epsilon$$

since $\epsilon$ is arbitrary, the theorem follows