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


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.




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?


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.

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    $\begingroup$ This is definitely the right idea. The challenge (which is worth accepting) is how to make these ideas rigorous. Somewhere you're probably going to have to use the fact that $f(x)$ is (uniformly) continuous on $[a,b]$. $\endgroup$ Commented Mar 23, 2016 at 23:34
  • $\begingroup$ @GregMartin Thank you. I will see if this can lead me to a proof. $\endgroup$
    – Arbuja
    Commented Mar 23, 2016 at 23:54
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    $\begingroup$ Indeed, some hypothesis on $f$ are still omitted and still sorely lacking. If one assumes that $f$ is Riemann integrable (and $f$ being continuous implies this), then one knows that its Riemann sums $$S(\sigma,f)=\sum_if(x_i)(x_{i+1}-x_i)$$ converge to its integral $I(f)$ when the mesh $\max\limits_i(x_{i+1}-x_i)$ of the subdivision $\sigma=(x_i)$ goes to zero. In your setting, $\sigma$ is obtained by throwing away some non consecutive points of the regular subdivision with step $\frac1n$, hence the mesh of $\sigma$ is $\frac2n$ and $S(\sigma,f)\to I(f)$. Remains to estimate ... $\endgroup$
    – Did
    Commented Jun 2, 2016 at 8:11
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    $\begingroup$ ... the difference between $S(\sigma,f)$ and the sums you consider, which is $J(\sigma,f)=(b-a)\frac1n\sum\limits_if(y_i)$, where $(y_i)$ are the points you throw away. These are themselves a subdivision of size $n/d$ hence $J(\sigma,f)\to\frac1dI(f)$ and finally, the limit you are after is $$\left(1-\frac1d\right)\cdot\int_a^bf(x)dx.$$ $\endgroup$
    – Did
    Commented Jun 2, 2016 at 8:17
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    $\begingroup$ What? $ $ $ $ $ $ $\endgroup$
    – Did
    Commented Sep 15, 2016 at 13:52

2 Answers 2


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.

  • $\begingroup$ Thank you for answer. I gave you the bounty since you gave true proof. I may accept this answer later on. $\endgroup$
    – Arbuja
    Commented Jun 2, 2016 at 20:19
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    $\begingroup$ Thank you. I keep editing the post to make it better. $\endgroup$
    – zhw.
    Commented Jun 3, 2016 at 20:55
  • $\begingroup$ Sorry, I meant to say that $f$ is integrable from $[a,b]$ but you gave me a rough idea of proving the problem. $\endgroup$
    – Arbuja
    Commented Jun 17, 2016 at 1:15
  • $\begingroup$ So I can take it from here $\endgroup$
    – Arbuja
    Commented Jun 17, 2016 at 1:26

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


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

since $\epsilon$ is arbitrary, the theorem follows

  • $\begingroup$ Thank you for your proof. I never had formal training in mathematics so since its rigorous it's hard to understand. I'll see what I can learn from it. $\endgroup$
    – Arbuja
    Commented Jun 1, 2016 at 0:56
  • $\begingroup$ @Arbuja since $f(x)$ is continuous you can replace fact 3 as: $f$ can be approximated by a step function $l_m$ at every point such that for every interval $|f-l_m| \le \epsilon$ $\endgroup$
    – ibnAbu
    Commented Jun 1, 2016 at 21:54
  • $\begingroup$ see this en.wikipedia.org/wiki/Step_function $\endgroup$
    – ibnAbu
    Commented Jun 1, 2016 at 21:56
  • $\begingroup$ I am not sure if this is a proper proof. The user called "Did" ,made comments below my post suggesting that the proof should include the mesh of reimmen sums. $\endgroup$
    – Arbuja
    Commented Jun 2, 2016 at 12:11

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