If $\int_{S_2}udS=0$ can we say that $u=0$ on $S_2$?

Question

Suppose that $\Omega\subset\mathbb{R}^2$. If we have a ball $B=B(x_0,\rho)\subset\Omega$ such that $\partial B=S_1\cup S_2,\;S_1\cap S_2=\emptyset,$ and $u=0$ on $S_1$ and $u\geq0$ on $S_2.$ Then, if $$\int_{S_2}udS=0$$ can we say that $u=0$ on $S_2$?

Answer 1

If $u \ge 0$ is continuous on a space with measure $(X, \mu)$, the open subset of $X$ are not null sets and $\int _X u \ \Bbb d \mu = 0$, then it follows that $u = 0$, so the answer to your question is affirmative.

Proof: assume that there exist $x_0 \in X$ with $u(x) > 0$. Let's say that $u(x_0) > a > 0$. By continuity, there must exist a whole neighbourhood $U$ of $x_0$ such that $u(x) > a$ on $U$. Notice that, if $\chi_U$ is the characteristic function of $U$, then $u \ge \chi_U u$, so by the fundamental properties of the integral you get

$$0 = \int _X u \ \Bbb d \mu \ge \int _X \chi_U u \ \Bbb d \mu = \int _U u \ \Bbb d \mu \ge \inf u \big|_U \mu (U) \ge a \mu (U) ,$$

whence $\mu (U) = 0$, which is not possible for the kind of measure that you are working with ($\Bbb d S$ being the Riemannian volume form of $S$).