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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 say that $\mathbb{R}^n\setminus E$ is simply connected.

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.

Let $E\subset \mathbb{R}^n$ be a (measurable) subset with $\mathcal{H}^{n-1}(E)=0$. 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 this paper), in which case the conclusion can be strengthened to say that $\mathbb{R}^n\setminus E$ is simply connected.

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 say that $\mathbb{R}^n\setminus E$ is simply connected.