Does approximately Fréchet differentiable imply approximately Gateaux differentiable? In what follows, $\mu^n$ denotes $n$-dimensional Lebesgue measure, and $B_r(x)$ is the open ball with radius $r$ centered at $x$.

In elementary calculus, if we have a function $f : \mathbb{R}^n \rightarrow \mathbb{R}^m$ which is Fréchet differentiable at a point $p \in \mathbb{R}^n$, then $f$ is automatically Gateaux differentiable at $p$ also.
However in geometric measure theory, we have the notion of approximate limits and differentiability. The definitions are valid as long as $f$ is a measurable function, and are given as follows:


*

*$f$ is approximately Fréchet differentiable at $p$ if there exists a linear map $L : \mathbb{R}^n \rightarrow \mathbb{R}^m$, such that for all $\varepsilon \in \mathbb{R}^{>0}$: $$\lim_{r \rightarrow 0^+} \frac{\mu^n(\{x \in B_r(p) : \frac{|f(x) - f(p) - L(x-p)|}{|x-p|} > \varepsilon \})}{\mu^n(B_r(p))} = 0 $$




*$f$ is approximately Gateaux differentiable at $p$ if for all unit vectors $u \in \mathbb{R}^n$ there exists $L_u \in \mathbb{R}^m$, such that for all $\varepsilon \in \mathbb{R}^{>0}$: $$\lim_{r \rightarrow 0^+} \frac{\mu^1(\{t \in (-r,r) : |\frac{f(p+tu) - f(p)}{t} - L_u| > \varepsilon \})}{2r} = 0 $$

Notice that here approximately Gateaux differentiable means that all the approximate directional derivatives exist, not just almost all of them. In fact it is this distinction which prevents Fréchet necessarily implying Gateaux pointwise.

Now my question is, what if we have everywhere approximately differentiable and not just pointwise? Specifically, if $f$ is everywhere approximately Fréchet differentiable, is $f$ necessarily approximately Gateaux differentiable?
For what it's worth, given any fixed unit vector $u$, it's known that the approximate $u$-directional derivative exists almost everywhere (see theorem 2 here https://encyclopediaofmath.org/wiki/Approximate_differentiability). My question is equivalent to asking if this can be extended to existing everywhere, once we have everywhere approximately Fréchet differentiable?
 A: I have an example which does not strictly answer the question, but a slightly weaker one.

Theorem 1. There is a function $f:\mathbb{R}^n\to\mathbb{R}$ that is approximately Fréchet differentiable almost everywhere, but $f$ is approximately Gateaux differentiable on a set of measure zero.

The construction is based on the existence of Nikodym sets:

Theorem 2. (Nikodym) There is a Borel set $N\subset\mathbb{R}^2$ such that $\mu^2(N)=0$ and for every $x\in\mathbb{R}^2$ there is a line $L$ though $x$ for which $L\setminus\{x\}\subset N$.

This is Theorem 11.7 in Mattila's "Fourier Analysis and Hausdorff Dimension".
Proof of Theorem 1. Let $A=\mathbb{R}^2\setminus N$ and let $f=\chi_A$ be the characteristic function of $A$. Since the complement of $A$ has measure zero, $f$ is approximately Fréchet differentiable at every point of $A$ with $Df=0$.
On the other hand if $x\in A$ and $L$ is the linie form Theorem 1, then the function $f$ restricted to that line equals $1$ at $x$ and zero everywhere else so it is not approximately differentiable in the direction of $L$ and hence $f$ is not approximately Gateaux differentiable at any point $x\in A$.   $\Box$
A: It's very possible that I'm misunderstanding something here (Update: I was indeed, the answer to my comment below the question clarified the issue), but to me it seems that the implication has no hope to be true.
Take a function $f:\mathbb{R}^2\to\mathbb{R}$ which is approximately Fréchet differentiable. Then take single direction $\nu$ and change $f$ with a new function $f^*$ only on the line $\{t\nu,\ t\in\mathbb{R}\}$ in such a way that the limit
$$\lim_{r \rightarrow 0^+} \frac{\mu^1(\{t \in (-r,r) : |\frac{f^*(p+t\nu) - f^*(p)}{t} - L| > \varepsilon \})}{2r} $$ isn't zero, or does not exist for any linear map $L$. You can take the intersection between the graph of $f^*$ and the plane determined by $\nu$ to be whatever ugly monster you want (say, the graph of the Dirichlet function). This cannot affect the measure of the set $\{x \in B_r(p) : \frac{|f^*(x) - f^*(p) - L(x-p)|}{|x-p|} > \varepsilon\}$, so $f^*$ is still approximately Fréchet differentiable... what am I missing?
