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...