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?