# Thread: a property of point of density

1. ## a property of point of density

I have a problem, that I can't do.

Let be $E$ a Lebesgue-measurable set of $R^n$, and let be $0$ a point of density (of Lebesgue) for $E$, i.e., for all set Lebesgue-measurable $B$ which contains $0$, we have
$\lim_{B\to0}\frac{|E\cap B|}{|B|}=1.$

Let be a sequence ${x_k}\rightarrow0$. Proof that
$\lim_{k\to\infty}\frac{D(x_k)}{||x_k||}=0,$
where $D(y)$ is the distance of $y$ from the set $E$.

Can someone help me? Thanks

( $|\cdot|$ denote the measure of Lebesgue.)

2. Originally Posted by miccoli
I have a problem, that I can't do.

Let be $E$ a Lebesgue-measurable set of $R^n$, and let be $0$ a point of density (of Lebesgue) for $E$, i.e., for all set Lebesgue-measurable $B$ which contains $0$, we have
$\lim_{B\to0}\frac{|E\cap B|}{|B|}=1.$

Let be a sequence ${x_k}\rightarrow0$. Proof that
$\lim_{k\to\infty}\frac{D(x_k)}{||x_k||}=0,$
where $D(y)$ is the distance of $y$ from the set $E$.

Can someone help me? Thanks

( $|\cdot|$ denote the measure of Lebesgue.)
Hi,

Let me denote by $B(x,r)$ the ball of center $x$ and radius $r$.

The main observation is that, for all $x$, since $d(x)\leq\|x\|$ (this is because $0\in E$), we have $B(0,2\|x\|)\cap E\subset B(0,2\|x\|)\setminus B(x,d(x))$. After some thought, this inclusion should become obvious.

Then you take the measure of the subsets: $|B(0,2\|x\|)\cap E|\leq |B(0,2\|x\|)|-|B(x,d(x))|$, you divide by $B(0,2\|x\|)$, and you should be able to conclude very shortly.

Don't hesitate telling us what you tried and what failed if you want more help.

3. Hello!

Thanks very much!! But I have a question: $0$ is a point of density for $E$, but this doesn't implicate that 0 belongs to $E$. If $0\notin E$, can I say $d(x)\leq||x||$ for all $x\in\mathbb{R}^n$?

Another question: What $|B(0,2||x_k||)|$ and $|B(x_k,d(x_k))|$ are equal to?

I have a question: $0$ is a point of density for $E$, but this doesn't implicate that 0 belongs to $E$. If $0\notin E$, can I say $d(x)\leq||x||$ for all $x\in\mathbb{R}^n$?
That's a good point. However, 0 must be a limit point of $E$. Otherwise, for $B$ small enough around 0, $B\cap E=\emptyset$ hence $\frac{|B\cap E|}{|B|}=0$, which is false. Thus it doesn't really matter. But the safest way is to consider $B(0,2\|x\|)\setminus B(x,\frac{d(x)}{2})$ !
Another question: What $|B(0,2||x_k||)|$ and $|B(x_k,d(x_k))|$ are equal to?
You have of course $|B(x,r)|=C_n r^n$ where $C_n$ is the volume of the unit ball, and the good news is you don't need the value of $C_n$, it should simplify.