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

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)

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

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

• $\frac{x-A}{dx}$ is an abuse of notation. It doesn’t mean anything. – Thomas Andrews Dec 3 '19 at 0:08
• 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 ? – More Anonymous Dec 3 '19 at 7:08