A unit speed differentiable curve which avoids a null set of directions For each $x \in \mathbb{R}^2$, let $N_x$ be a null set of the unit circle $S^1$ (with respect to $1$-dimensional Lebesgue measure).
Further given that for each $u \in S^1$, we have that $u \not \in N_x$ for almost all $x \in \mathbb{R}^2$.
Does there necessarily exist a unit speed differentiable curve $\gamma : (-\varepsilon, \varepsilon) \rightarrow \mathbb{R}^2$ such that:

*

*$\gamma'(t) \not \in N_{\gamma(t)}$ for all $t \in (-\varepsilon,\varepsilon)$?

 A: No.
Here is an example where such curve does not exist.
Let
$$N_x = \Bigl\{(u_1,u_2) \in S^1: u_1 = (\pi+q_1)x_1 + q_2\\
\text{or } u_2 = (\pi+q_1)x_2 + q_2;\; q_1,q_2\in \mathbb{Q}\Bigr\}. \tag{1}$$
Then $N_x$ is countable and for each $u\in S^1$
we have $u\in N_x$ only if $x_1$ or $x_2$ takes one of countably many values. Hence, both $N_x$ and $\{x: u\in N_x\}$ are null sets.
Yet no such $\gamma$ exists. In order to do this, we first show the following lemma.
Lemma 1. Let $g\colon (a,b) \to \mathbb{R}$ be a differentiable function satisfying $\forall t \in (a,b)\;\; \dot g(t) \notin \mathbb{Q}$. Then $g$ is affine, i.e. $g(t) = \alpha + \beta t$ for some $\alpha,\beta\in\mathbb{R}$.
Proof. $g$ is continuous because it is differentiable. Consider the set
$$S = \left\{\frac{g(c_2) - g(c_1)}{c_2 - c_1}: c_1,c_2 \in (a,b), c_1 < c_2\right\}. \tag{2}$$
As an image of a connected set under a continuous map, $S$ is a connected subset of $\mathbb{R}$. On the other hand, from Mean value theorem $S\cap \mathbb{Q} = \emptyset$. Therefore, $S$ consists of a single point. If we denote that point by $\beta$, $t\mapsto g(t)- \beta t$ is a constant function (from the definition of $S$). $\;\square$
Now, assume to the contrary that such curve $\gamma$ exists. For any $q\in\mathbb{Q}, i \in \{0,1\}$ we can apply the lemma to the function $g_q: t \mapsto \gamma_i(t) - (\pi + q)\int_{0}^t \gamma_i(t') dt'$ to see that the function $\dot g_q: t \mapsto \dot\gamma_i(t) - (\pi + q_1) \gamma_i(t)$ is constant. Since $\gamma_i(t) = g_0(t) - g_1(t)$, we see that $\gamma_i$ is constant, contradicting the assumption of unit speed of $\gamma$.
A: Step 1 : We will define $N_o$ where $o=(0,0)$ is the origin in $
 \mathbb{R}^2$ : $$N_o =
\bigg\{ \bigg(\cos\ t,\sin\ t \bigg) \in \mathbb{S}^1\bigg|
t=\frac{2\pi}{2^i}k,\ i,\ k\in \mathbb{Z}_+\bigg\}
$$ where
$\mathbb{Z}_+$ is a set of nonnegative integers.
For convenience, we call a real number of type $\frac{2\pi}{2^i}k$ a
rational type. If not, we call it an irrational type.
Assume that $p$ is a point in $\mathbb{R}$ and we define its Euclidean distance $r= |p|$.
If $r$ is rational type, then $R_r$ is $r$-rotation map on
$\mathbb{S}^1$ with a clockwise direction. If $r$ is irrational,
then $R_r$ is $r$-rotation map with a counter-clockwise direction.
We define $T_r$ to be a $r$-rotation map on $\mathbb{S}^1$ with a
counter-clockwise direction for any $r>0$.
Then we define a null set at each point $p$ :
$$ N_p = R_r(N_o )
$$
Step 2 : Assume that there is a unit speed curve $c$ s.t.
a continuous tangent vector $c'(t)$ avoids the given null sets
$N_{c(t)}$ for $0\leq t\leq \epsilon$. We will prove that there is
no such curve $c$.
By a construction of the null set $N_{c(t)}$ we have uniquely
$v(t)\in \mathbb{S}^1 - N_o$ s.t.
$$R_{|c(t)|}(v(t) )=c'(t) = T_{|c(t)|} (w(t))  $$
Define $$A = \bigg\{ v(t) \in N_o\bigg| |c(t)| \ {\rm is\ rational}
\bigg\},\ B =\bigg\{v(t)\in N_o\bigg|
|c(t)|\ {\rm is\ irrational}\bigg\}$$ and $C=\{w(t)\in N_o \}$.
Clearly, $B$ is in $C$. For $v(t)=(\cos\ \phi,\sin\ \phi)\in A$, we
have $$c'(t)=\bigg(\cos (\phi -|c(t)| ),\sin (\phi -|c(t)| ) \bigg)
$$ and define
$$ A'=\bigg\{ \bigg(\cos\ (\phi -2|c(t)|),\sin\ (\phi -2 |c(t) | ) \bigg) \bigg\}$$
Then $A'\cup B=C$ is a connected compact arc in $\mathbb{S}^1$.
Hence for some $t$, $ \phi -2|c(t)| $ should has a rational type
which contradicts to the fact that $\phi$ is irrational and $|c(t)|$
is rational. Hence we conclude that $C$ is a point set.
Case 1 - $|c(t)|$ is const : Hence $c$ goes along a circular
arc s.t. the number of its tangent vectors is infinite, which is
impossible.
Case 2 - $|c(t)|$ is not const : Then $A =\{ \theta_A\},\
B=\{\theta_B\}$ are point sets.
Assume that $0<|c(t_1)| -| c(t_2)|$ is small. If rational type $r$
and irrational type $ r'$ are arbitrarily close to $|c(t_1)|$, then
the vectors $e^{i(\theta_A-r)},\ e^{i(\theta_B+r)}$ are arbitrarily
close to the tangent vector $c'(t_1)$. Hence $\theta_A-\theta_B
=2|c(t_1)| $ up to $\pi$. Hence we have $2|c(t_1)|=2|c(t_2)|$ up to
$\pi$, which is impossible. Hence we complete the proof.
