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

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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:

> 1. $f$ is approximately Fréchet differentiable at $p$ if there exists a linear map $L : \mathbb{R}^n \rightarrow \mathbb{R}^m$, such that for all $\varepsilon \in \mathbb{R}^{>0}$: $$\lim_{r \rightarrow 0^+} \frac{\mu^n(\{x \in B_r(p) : \frac{|f(x) - f(p) - L(x-p)|}{|x-p|} > \varepsilon \})}{\mu^n(B_r(p))} = 0 $$

> 2. $f$ 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.

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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).