# Prove the following limits of a sequence of sets?

(See the paper in the attatchment for a background behind the problems.)

Suppose $A=\mathbb{Q}$ where $f:A\to\mathbb{R}$ such that:

$f(x)= \begin{cases} 1 & x\in\left\{{(2n+1)}/{2m}:n\in\mathbb{Z}, m\in\mathbb{N}\right\}\\ 0 & x\not\in\left\{{(2n+1)}/{2m}:n\in\mathbb{Z}, m\in\mathbb{N}\right\}\\ \end{cases}$

and, in sequel, the integral of $f$ is taken w.r.t to the counting measure.

1. If $(F_r)_{r\in\mathbb{N}}= \left(\left\{c/r!:c\in\mathbb{Z},\, -r\cdot r!\le c\le r\cdot r! \right\}\right)_{r\in\mathbb{N}}$ prove that:

$$\forall(\epsilon>0)\exists(N\in\mathbb{N})\forall(r\in\mathbb{N})\left(r\ge N\Rightarrow\left|\frac{1}{\left|F_r\right|}\int_{F_r}f(x)\, d\mathbf{x}-1\right|< \epsilon\right)$$

2. If $(F_r)_{r\in\mathbb{N}}= \left(\left\{c/d:c\in\mathbb{Z},\, d\in\mathbb{N},\, d\le r,\, -dr\le c\le dr \right\}\right)_{r\in\mathbb{N}}$ prove that:

$$\forall(\epsilon>0)\exists(N\in\mathbb{N})\forall(r\in\mathbb{N})\left(r\ge N\Rightarrow\left|\frac{1}{\left|F_r\right|}\int_{F_r}f(x)\, d\mathbf{x}-1/3\right|< \epsilon\right)$$

• Paul F
+1

Low bounty!

• How much should it be?

• Well, this looks like a tricky question to me. I think it should be at least doubled or tippled. Think about how long it would take someone to answer, it would give you an idea.

• I will double it. I think for a highly skilled mathematician this should be an exercise.

• Yes, but highly skilled mathematicians also tend to value their time. I'm busy at the moment, but I may have some time over the weekend to think about it, if no one else answers by then.