Approximate derivative

Remarks

Example: let $f(1/n) = 1$ for $n \ge 2$ and $f(y) = 0$ at all other $y \in (-1, 1)$. At $x = 0$ the ordinary derivative does not exist, since the quotient at $y = 1/n$ equals $n$. The set $E = (-1, 1) \setminus \set{1/n : n \ge 2}$ has density $1$ at $0$, because the removed set is countable, and on $E$ the quotient is $0$; so $f'_{\mathrm{ap}}(0) = 0$.

Non-example: let $f(y) = 1$ for $y \ge 0$ and $f(y) = 0$ for $y < 0$, at $x = 0$. For $y < 0$ the quotient is $-1/y$, which tends to $+\infty$ as $y \to 0^-$. Every set of density $1$ at $0$ contains points $y < 0$ arbitrarily close to $0$. So $f$ has no approximate derivative at $0$.

If $f'(x)$ exists, then $f'_{\mathrm{ap}}(x)$ exists and equals it, by taking $E = I$. The approximate derivative is unique when it exists, because two sets of density $1$ at $x$ meet in a set of density $1$ at $x$.