Proof step in Rademacher's Theorem

Question

In the proof of Rademacher's theorem, we assume that $f: \Bbb R^n \to \Bbb R$ is a Lipschitz function and $v \in \Bbb R^n$ is a vector with $\Vert v \Vert = 1$. Our aim is to show, that $$ \mathrm D_v f(x) = v \cdot \text{grad} f(x) \; ,$$ where $\mathrm D_v f(x)$ is the directional derivative of $f$ at the point $x$ in the direction $v$, and on the right sight of the equation we have the euclidean scalar product of $v$ and the gradient of $f$ at the point $x$. In the proof we have shown, that $$ \int_{\Bbb R^n} \mathrm D_v f(x) \zeta(x) \; \mathrm dx = \int_{\Bbb R^n} [v \cdot \text{grad}f(x)] \zeta(x) \; \mathrm dx $$ for any $\zeta \in C_c^\infty(\Bbb R^n)$. Why is this enough to know, that $$ \mathrm D_v f(x) = v \cdot \text{grad} f(x) $$ holds?

Answer 1

A start: Take the contrapositive and suppose $f$ and $g$ were well-behaved functions that were not equal (what is a sufficient criterion for well-behavedness?). You could figure then there would be some $\zeta$ for which the integral equality would not hold, for example, if $\zeta$ were some kind of smooth bump function whose bump was in the neighborhood around some point at which $f$ and $g$ are not equal.