Suppose that $M^2$ is a closed Riemannian manifold and that $u$ is a $C^2(M\setminus S),$ where $S$ is a closed measure set consisting possibly on a enumerable amount of points. Can we still conclude that $\int\_M\Delta u =0?$



Is $\int_M\Delta u = 0$ if $u$ is not $C^2$ on a set of measure zero?
You have to say what is the meaning of $\Delta u$, and of $\int\Delta u$. For the integral to have a meaning, $u$ has to be a distribution and $\Delta u$ has to be a (signed, Radon) measure. Such distributions are called $\delta$-subharmonic functions. 


Then $\int\_M\Delta u=0$
without any assumptions about $C^2$ or $S$. (This is called the "Main Theorem" about compact Riemann surfaces
in S. Donaldson, Riemann surfaces, Chap 8, Thm. 5).


But if you understand $\Delta u$ in the naive, classical sense for $C^2$ function, and understand your integral as $\int\_{M\backslash S}\Delta u$ then of course this does not have to be $0$.


For example, on the Riemann sphere, let $u(z)=0,\,|z|<1,\; \; u(z)=|z|,\, 1<z<2,\;\; u(z)=2,\, |z|>1$. This function is $C^2$ except on the set of measure zero
consisting of two circles $|z|=1$ and $|z|=2$. But
$$\left(\int\_{|z|<1}+\int\_{1<|z|<2}+\int\_{|z|>1}\right)u(z)>0.$$


Even simpler example is $u(z)=|z|$ on the Riemann sphere, $\Delta u>0$ when
restricted to $0<|z|<\infty$ and $\int\_{0<|z|<\infty}u(z)=+\infty$.


