# Can we always extract a proper Hausdorff measurable subset from a Hausdorff measurable set?

I also put this question on MSE here

Let $\Gamma\subset \Omega\subset \mathbb R^N$ be such that $\mathcal H^{N-1}(\Gamma)<+\infty$ (this also implise that $\Gamma$ is Hausdorff measurable).

Let $\mathcal S^{N-1}$ be the unit sphere in $\mathbb R^N$ and let $\nu\in \mathcal S^{N-1}$ be a fixed direction. We set \begin{align}\label{slicing_notation} \begin{cases} \pi_\nu: = \{x\in\mathbb R^N:\,\left<x,\nu\right>=0\};\\ \Omega_{x,\nu}:=\{t\in\mathbb R:\, x+t\nu\in\Omega\}\,\,\text{ for }x\in\pi_\nu;\\ \Omega_\nu: = \{x\in\pi_\nu:\,\Omega_{x,\nu}\neq \varnothing\}. \end{cases} \end{align} That is, the sets $\Omega_{x,\nu}$ are the one-dimensional slices of $\Omega$ indexed by $x\in\pi_\nu$.

We assume that $\Gamma$ has property such that $$\mathcal H^0(\Omega_{x,\nu}\cap\Gamma)\geq 2$$ for each $x\in\Omega_\nu$

My question is: Would it be possible to extract a subset $\Gamma'$ from $\Gamma$ such that $\Gamma'$ is $\mathcal H^{N-1}$ measurable and satisfies $$\mathcal H^0(\Omega_{x,\nu}\cap\Gamma')= 1 \tag 1$$ for each $x\in\Omega_\nu$?

To satisfy $(1)$ is easy, we just need to choose one point from each set $\Omega_{x,\nu}\cap\Gamma$ for each $x\in\Omega_\nu$ to form $\Gamma'$. However, I found it is hard to make a good choice so that $\Gamma'$ is $\mathcal H^{N-1}$ measurable...