I also put this question on MSE [here][1]

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... 

Please advise!


  [1]: http://math.stackexchange.com/questions/1836255/can-we-always-extract-a-proper-hausdorff-measurable-subset-from-a-hausdorff-meas