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We say a real-valued function $f$ on $\textbf{R}^n$ has an approximate differential at $x_0$ if there is a measurable set $H \subset \textbf{R}^n$ containing $0$ and having $0$ as a point of Lebesgue density, such that:

$f(x_0+h) = f(x_0) + A.h + o(|h|)$ for $h \in H$, $h \rightarrow 0$ and a vector $A \in \textbf{R}^n$

How do we show the approximate differential $A$ is unique here? We'd have to show that if $H' \subset \textbf{R}^n$ was another measurable set containing $0$ and having $0$ as a point of Lebesgue density, and there was another vector $A'$ such that $f(x_0 + h') = f(x_0) + A'.h' + o(|h'|)$ for $h' \in H'$, $h' \rightarrow 0$, then $A=A'$.

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    Consider $h=h'$. – 2017-01-12
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    Of course. Both sets have $0$ as a point of density, we can pick a sequence going to $0$ where $h = h'$. Thanks! – 2017-01-12

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