Let $X$ be a connected Riemann surface and let $f,g$ two complex-valued functions on $X$ such that $f$ is locally integrable and $g$ is $C^\infty$ and compactly supported. Why do we have the following equality? $$\int_X f \partial\bar\partial g=\int_X g \partial\bar\partial f$$
Here $\partial$ and $\bar\partial$ are the usual differential operators on $(p,q)$-forms. Clearly functions are $(0,0)$-forms.
Because $g$ is compactly supported, we can apply Stokes's Theorem to a compact Riemann surface $Y\subset X$ with boundary so that $g=0$ on a neighborhood of $\partial Y$. Then there will be no boundary terms. Note that \begin{align*} d(f\bar\partial g) &= fd\bar\partial g + df\wedge \bar\partial g = f\partial\bar\partial g + df\wedge\bar\partial g, \quad\text{and} \\ d(g\partial f) &= gd\partial f + dg \wedge\partial f = g\bar\partial\partial f - \partial f\wedge dg. \end{align*} Therefore (remembering that there are no $(2,0)$- or $(0,2)$-forms on a Riemann surface) $$\int_X f\partial\bar\partial g = -\int_X df\wedge\bar\partial g = -\int_X \partial f\wedge dg = \int_X dg\wedge\partial f = -\int_X g\bar\partial\partial f = \int_X g\partial\bar\partial f.$$