If $\mathcal{H}^{n-1}(E)=0$ then $\mathbb{R}^n\setminus E$ is connected

Question

Let $E\subset \mathbb{R}^n$ be a (measurable) subset with $\mathcal{H}^{n-1}(E)=0$, where $\mathcal H^{n - 1}$ is the ($n - 1$)-dimensional Hausdorff measure. I want to know if $\mathbb{R}^n\setminus E$ must be connected (even path-connected?).

I am only aware of a proof in the case when $\mathcal{H}^{n-2}(E)=0$ (which can be found in the appendix of Simon and Wickramasekera - A Frequency Function and Singular Set Bounds for Branched Minimal Immersions), in which case the conclusion can be strengthened to saying that $\mathbb{R}^n\setminus E$ is simply connected.

Answer 1

Yes, $\mathbb{R}^n\setminus E$ has to be path-connected.

Let $x,y\in\mathbb{R}^n\setminus E$, we will prove that there are many paths going from $x$ to $y$ inside $\mathbb{R}^n\setminus E$. We can suppose WLOG that $x=(1,0,\dots,0)$ and $y=(-1,0,\dots,0)$.

For each $p$ in $\{0\}\times\mathbb{R}^{n-1}$ you can consider the path $\gamma_p:[0,1]\to\mathbb{R}^n$ going from $x$ to $y$ and passing through $p$ given by $\gamma_p(t)=(\cos(\pi t),0,\dots,0)+p\sin(\pi t)$ (an arc of ellipse).

Now, letting $\Gamma:=\bigcup_p\gamma_p((0,1))=(-1,1)\times\mathbb{R}^{n-1}$, we can define a `projection' $\pi:\Gamma\to\{0\}\times\mathbb{R}^{n-1}$ defined by $\pi(\gamma_p(t))=p$ for all $t\in(0,1)$ and for all $p$ (this is well defined because the paths $\gamma_p$ are disjoint except in the extreme points $x,y$).

As $\pi$ is a locally Lipschitz map and $H^{n-1}(E\cap\Gamma)=0$, $H^{n-1}(\pi(E\cap\Gamma))$ is also $0$. Thus $\pi(E\cap\Gamma)$ is not the entire $\{0\}\times\mathbb{R}^{n-1}$, and letting $p\in\{0\}\times\mathbb{R}^{n-1}\setminus\pi(E\cap\Gamma)$, the path $\gamma_p$ is a path from $x$ to $y$ inside $\mathbb{R}^n\setminus E$.