Lebesgue measure of a set

Question

Question is :

Let $K\subseteq \mathbb{R}^n$ is compact then $M=\{x\in \mathbb{R}^n : d(x,K)=1\}$ is of measure zero..

I did the following for $n=1$..

As $K$ is closed, $\mathbb{R}\setminus K$ is open so we have $\mathbb{R}\setminus K=\bigcup_{i=1}^{\infty}(a_i,b_i)$

As $K$ is closed no element of $K$ can be at a positive distance from $K$.

So, $M\subset \mathbb{R}\setminus K$.. Now, each interval $(a_i,b_i)$ can have atmost one elemnet whose distance from $K$ is $1$... I assume that each of these intervals can be choosen to be of length $<1$..

I am not sure whether this is true..

I did not use that $K$ is compact.. I just used that $K$ is closed here..

For general cases also i can think of some simple compact sets in $R^2$... Unit disks for which $M$ would then be a circle in $\mathbb{R}^2$ which is clearly of measure $1$..

I can not think of anything more than this..

Help me to understand this better...

Answer 1

Now, each interval $(a_i,b_i)$ can have atmost one elemnet whose distance from $K$ is $1$.

Why is that? Are the intervals assumed to be mutually disjoint? Then this is definitely wrong. Consider $K = [0,1]\cup [4,5]$. Then the interval $(1,4)$ contains the two points $2$ and $3$ whose distance from $K$ is $1$. However, if you replace "one element" by "two elements" above, you are fine and so is your argument. This even proves that $M$ is a finite set in the case of $\mathbb R$. BTW, you don't need to assume that the length of each interval is smaller than one. For the general case: The set $M$ is the boundary of the open set $K + B_1$, where $B_1$ is the open unit ball. Maybe this helps.

Answer 2

OK, first if $K$ is a disk in $\mathbb{R}^2$, then $M$ is a circle, whose measure in $\mathbb{R}^2$ is zero. Also, I believe $M$ is closed because it is the inverse image of a closed set, $\{1\}$, under a continuous function, $d(\cdot,K)$.

Here's a proof sort of like yours for dimension 1. In $\mathbb{R}$ for any $x\in M$ we have either $x+1\in K$ or $x-1\in K$ or both, and in either case $(x-1,x+1)\cap K=\emptyset$. We want to show that each element of $M$ is isolated, i.e., for every $x\in M$ there is an interval around $x$ where $x$ is the sole element of $M$. That will show that $M$ is finite because $K$ is bounded, $M\subset [\inf K-1,\sup K+1]$, and if it were infinite it would have to have a point of accumulation.

If $x+1\in K$ then $M\cap (x,x+1)=\emptyset$. If also $x-1\in K$, then $M\cap (x,x+1)=\emptyset$ because for $y\in(x-1,x)$ we have $d(y,k)=y-(x-1)<1$. Thus, if $x-1\in K$ also, $x$ is isolated.

Assume then $x+1\in K$ and $x-1\notin K$. Then, because $K$ is closed, $(x-1-\epsilon,x-1]$ doesn't intersect $K$ for some $\epsilon>0$. Thus, any $y\in M$ which is less than $x$ must be $\le x-\epsilon$, and so $x$ is isolated, as desired.

The case $x-1\in K$ is similar.