5
$\begingroup$

Mathematicians like to ask covering various sets with open intervals and the answers to these riddles have strange tendencies to become strange lemma's or theorems Heine-Borel Theorem, lebesgue's Number Lemma, Vitali Covering Lemma, Besicovitch Covering Theorem, (etc)

-3blue1brown

By open interval he means the stretch of real numbers: $a<x<b$.

My challenge is to strictly weighted-ly cover all the rational numbers between $A$ and $B$ with open intervals. By weighted-ly cover means each paticular rational number lies in a particular interval. However, each interval has a weight $a_i$. The challenge here is that weighted sum must be less than $k'$.

Before we proceed let's return to normal measure theory where $a_i=1$ for all weights. Then we have $k' = \int_{A}^B dx= B-A$.

Now, let us look at limit of a sum as Riemann integral:

$$ {\displaystyle \Delta x={\frac {B-A}{n}}}$$

$$ A + r \Delta x \to x$$

OR

$$ r \to \frac{x -A}{\Delta x}$$

Note: $r$ is a dummy variable.

Now, in the limit $\Delta x \to 0$

$$ \int_A^B f(x) dx = \lim_{\Delta x \to 0}\Delta x\left[f(a+\Delta x)+f(a+2\,\Delta x)+\cdots +f(b-\Delta x)\right]= \lim_{\Delta x \to 0}\sum_{r=1}^n f(a+ r \Delta x) \Delta x$$

So what will $k$ be in the case where all $a_i$ aren't the same? Let's put this question in light of the answer:

Claim: If $\lim_{n \to \infty} \frac{\log^2(n)}{n}\sum_{r=1}^n |b_r| = 0$ and $f$ is smooth, then $$\lim_{k \to \infty} \lim_{n \to \infty} \sum_{r=1}^n a_rf\left(\frac{kr}{n}\right)\frac{k}{n} = \left(\lim_{s \to 1} \frac{1}{\zeta(s)}\sum_{r=1}^\infty \frac{a_s}{r^s}\right)\int_0^\infty f(x)dx.$$

where $a_r = \sum_{e|r} b_e$

All we now do is add the notation:

$$ \int_{A}^B a_{\frac{x-A}{dx} } f(x) dx= \lim_{k \to B-A} \lim_{n \to \infty} \sum_{r=1}^n a_rf\left(A+\frac{kr}{n}\right)\frac{k}{n}$$

In fact using the co-ordinate transformation $dy = f(x) dx$ with a function. We also define a coordinate transformation or mapping $ g(x) = y$. Hence,

$$ r \to \frac{x-A}{dy} \frac{dy}{dx}= f(x) \frac{x-A}{dy} = f(g^{-1} (y)) \Big( \frac{g^{-1} (y)- A}{dy} \Big) $$

This enables us to talk about coordinate transformations.

Question

What is the measure of $\int_{A}^B a_{\frac{x-A}{dx} } f(x) dx$ ?

If $$\lim_{n \to \infty} \frac{\log^2(n)}{n}\sum_{r=1}^n |b_r| = 0$$ and $$a_r = \sum_{e|r} b_e$$

$\endgroup$
  • 1
    $\begingroup$ $\frac{x-A}{dx}$ is an abuse of notation. It doesn’t mean anything. $\endgroup$ – Thomas Andrews Dec 3 '19 at 0:08
  • $\begingroup$ I'd say it's an extension of notation. Either way the question remains what is the measure of the lhs in the 2nd quotation ? $\endgroup$ – More Anonymous Dec 3 '19 at 7:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.