$\int_{\partial \Omega}fg=0$ iff $f=0$ ae, on Lipschitz domain Let $\Omega \subset \mathbb{R}^n$ be a bounded Lipschitz domain. Then for all $f \in L^1(\partial \Omega)$:
$\int_{\partial \Omega} fg=0, \; \forall g \in C^{\infty}(\partial \Omega) \Leftrightarrow f=0$ a.e. with respect to the measure of $\partial \Omega$.
In my opinion, "$\Leftarrow$" is easily shown since we have two bounded functions on the negligible set $supp_{\partial \Omega}(f)$, which implies that $\left\lvert \int_{\partial \Omega} fg\right\rvert \leq C \cdot \left\lvert supp_{\partial \Omega}(f)\right\rvert = 0$.
I'm struggling to show the second part though. It would be nice to have some steps of "smallest things" I have to show, please.
Also, the fact that $\Omega$ is bounded makes its boundary compact, which is obviously nice since we're dealing with integrals on it. But in which way does this matter to have a Lipschitz domain?
 A: One purpose of the Lipschitz condition is to ensure that there is some reasonable surface measure on $\partial \Omega$. In fact, I don't see another.
I presume that $C^{\infty}(\partial \Omega)$ are the restrictions of the functions from $C^{\infty}(U)$, where $U$ is some open subset of $\mathbb{R}^d$ containing $\partial \Omega$.
Since $\partial \Omega$ is compact, the mapping
$$T(g) = \int_{\partial \Omega} fg$$
is a continuous linear functional on $C(\partial \Omega)$. If we show that this mapping equals zero for all $g\in C(\partial \Omega)$, then by the uniqueness in the Riesz representation theorem we find that $f=0$ almost everywhere.
Let $g\in C(\partial \Omega)$. By Tietze's theorem, there exists a continuous extension of $g$ to the whole of $\mathbb{R}^d$, call it $\tilde{g}$. We may additionally assume that it is compactly supported. Then note that we may approximate $\tilde{g}$ by a smooth function in the supremum norm (by convolving with a compactly supported mollifier), that is, for every $\epsilon>0$ there exists a smooth compactly supported $f_\epsilon$ such that
$$\|f_\epsilon - g\|_{\infty} < \epsilon.$$
Since by assumption $T(f_\epsilon) = 0$, we get that
$$|T(g)| = |T(g-f_\epsilon)| \leq \|f_\epsilon - g\|_{\infty}\int |f| < \epsilon \int |f|,$$
so by letting $\epsilon\to 0$ we obtain $T(g) = 0$.
