Is $\iint \dfrac{1}{z} dxdy\neq 0$? I am trying to solve an exercise and at some point I came accross the integral
$$\iint_L \dfrac{1}{z} dxdy,$$
($z=x+iy$) where $L\subset \mathbb{C}$ is a compact set with positive two-dimensional Lebesgue measure. If this integral is not zero, then the excercise is solved. But is this true? Note that $\dfrac{1}{z}$ is locally integrable.
 A: The integral
$$\int_L \frac{1}{z}\,dx\,dy$$
may or may not vanish, that depends on $L$. For most $L$, the integral will be nonzero, but for example if $L = -L$, then by symmetry the integral vanishes.
As mentioned in the comments, the true purpose of the question is to show that
$$g(z) := \int_L \frac{1}{z-w}\,dm(w)$$
does not vanish identically.
This can be shown using standard estimates. For large $\lvert z\rvert$, consider
$$\frac{m(L)}{z} - g(z).$$
Say $L \subset D_R(0)$, and $\lvert z\rvert > 2R$. Then we have $\lvert z-w\rvert > \frac{1}{2} \lvert z\rvert$ for $w \in L$, and so
\begin{align}
\biggl\lvert \frac{m(L)}{z} - g(z)\biggr\rvert
&= \biggl\lvert \int_L \frac{1}{z} - \frac{1}{z-w}\,dm(w)\biggr\rvert\\
&\leqslant \int_L \biggl\lvert \frac{1}{z} - \frac{1}{z-w}\biggr\rvert\,dm(w)\\
&= \int_L \frac{\lvert w\rvert}{\lvert z\rvert\,\lvert z-w\rvert}\,dm(w)\\
&\leqslant \frac{2R}{\lvert z\rvert^2}m(L)\\
&< \frac{m(L)}{\lvert z\rvert},
\end{align}
which implies $g(z) \neq 0$ for $\lvert z\rvert > 2R$.
