Measure theory problem involving metric density function $\rho(E, x)$

Question

Working on the real line $(\mathbb{R})$, let $\mu : \mathscr{M} \rightarrow [0, +\infty]$ represent the Lebesgue measure ($\mathscr{M}$ is the set of measurable subsets of $\mathbb{R}$).

For $E \in \mathscr{M}$ and $x \in \mathbb{R}$, we define $$\rho(E, x) = \lim_{\delta \to 0+} \frac{\mu(E \cap (x - \delta, x + \delta))}{2\delta},$$ if the limit exists. The above limit is called the metric density of $E$ at $x$.

$(1)$ Given $E=(1,2)\cup(2,5]\cup\{6\}$, find the metric density of $E$ for all $x \in E$.

$(2)$ Let $\alpha \in (0, 1)$. Construct a set $E\subset\mathbb{R}$ such that $\rho(E, 0) = \alpha$.

My work and thoughts:

$(1)$ Since $x$ can be any real number, the intersection $E \cap (x - \delta, x + \delta)$ can be empty.

Also, if $(x - \delta, x + \delta) \subset E$, as $\delta \rightarrow 0+$ the intersection $E \cap (x - \delta, x + \delta) = x$.

In either case the measure equals zero. Is this correct?

For $(2)$ I have no idea.

For (ii), see Theorem 3.1 (proved on pp. 19-22) in Nathaniel Fizzel Grafton Martin's Ph.D. dissertation Metric Density of Sets (1959, Iowa State University). If you follow the details there, you should probably cite this reference in your solution. — Dave L. Renfro, Mar 03 '16 at 16:06
For (ii), see also Lebesgue measurable subset of $\mathbb{R}$ with given metric density at zero. — Dave L. Renfro, Mar 03 '16 at 16:30
@DaveL.Renfro I tried to understand both the theorem and example you provided but I must admit they are too complicated for my level right now. I will definitely come back to them when I have a better understanding of the subject. Thank you for your time anyway, much appreciated. — glpsx, Mar 03 '16 at 20:00

score 1 · Answer 1 · 2016-03-03T14:06:48.090

1

Let $x=\frac{3}{2}$ then, for small $\delta$ we have $$E\cap \left(\frac{3}{2}-\delta,\frac{3}{2}+\delta\right)=\left(\frac{3}{2}-\delta,\frac{3}{2}+\delta\right)$$

So, $\mu(E\cap(x-\delta,x+\delta))=2\delta$..

So??

edited Mar 03 '16 at 14:06

answered Mar 03 '16 at 13:57

As $\delta \rightarrow 0+$ the limit tends to zero. What I don't understand is why $E\cap (\frac{1}{2}-\delta,\frac{1}{2}+\delta)=\left(\frac{1}{2}-\delta,\frac{1}{2}+ \delta \right)$. The intersection should be equal to zero in my opinion since $\frac{1}{2} \notin E$. Could you explain this? – glpsx Mar 03 '16 at 14:05
@VonKar : CHeck the edit... I mean $\frac{3}{2}$... – Mar 03 '16 at 14:07
$\rho(E,x)=0$, if $x<1$; $\rho(E,x)=1/2$, if $x=1$; $\rho(E,x)=1$, if $1<x<5$; $\rho(E,x)=1/2$, if $x=5$ and $\rho(E,x)=0$, if $x>5$ – George Mar 04 '16 at 16:26
@GogiPantsulaia : What is your point... You can write it as answer... – Mar 04 '16 at 16:53

score 1 · Accepted Answer · answered Jun 12 '16 at 06:04

1

For Q(1). For $x\in E, $ if $x\in (1,2)\cup (2,5)$ then for all sufficiently small $d>0$ we have $$E\cap (x-d,x+d)= (x-d,x+d)$$ and $$(2 d)^{-1}\mu (E\cap (x-d,x+d))=(2 d)^{-1}\mu ((x-d,x+d))=(2 d)^{-1}\cdot 2 d=1.$$ You should be able to to show that $\rho (E,5)=1/2$ and $\rho (E,6)=0.$

answered Jun 12 '16 at 06:04

DanielWainfleet

59,529

Thank you very much! I was able to show that $\rho(E, 5) = 1/2$. Indeed, for $x = 5$, $E \cap (x -\delta,x + \delta)= (x - \delta)$ and $\mu(x - \delta) = \delta$. Therefore, $\rho(E, 5) = \lim_{\delta \to 0+} \delta / 2\delta = 1/2$. The case where $x = 6$ is the simplest. – glpsx Jun 13 '16 at 18:57

Measure theory problem involving metric density function $\rho(E, x)$

2 Answers2