In this article (Theorem 1), Besicovitch proved the following generalization of the well-known Cauchy-Goursat-Theorem
If a function $f(z)$ of a complex variable defined in an open simply connected domain $D$ is known to be bounded in the domain and to be differentiable (i.e. to have a finite derivative) at all points of the domain except possibly at the points of a set $E$ of linear measure zero, then ... $f(z)$ is holomorphic in the domain D.
The proof is similar to the classical proof of Goursat, but I don't really understand his geometric construction. Therefore, I hope to find someone who can help me work myself through the details.
He proves the statement by showing that the integral along any square contained in $D$ is zero (using Morera's Theorem) by contradiction, assuming that there is some square $A_1$ in $D$ of side $a$ such that \begin{align*} \left|\int_{A_1}f(z)dz\right|=m_1>0. \end{align*} Let $|f(z)|\leq M$ for all $z\in D$. Now he covers the set $E$ by a countable set $U$ of convex sets, such that \begin{align*} \sum_{B\in U}\text{diam } B<\frac{m_1}{64M}. \end{align*} Now he partitions $U$ into $U_1,U_2,U_3,\ldots,$ where $U_1$ contains all sets $B\in U$ with $\text{diam }B\geq a$, $U_2$ contains all sets $B\in U$ with $a/2\leq\text{diam }B<a$, $U_3$ contains all sets $B$ of $U$ with $a/4\leq\text{diam }B<a/2$ and so on.
Now comes the part I cannot follow. This is what he writes.
We now divide $A_1$ into 4 equal squares $A_{11}, A_{12}, A_{13}, A_{14}$ by parallels to the sides of $A_1$. To each area of $U_1$ there corresponds the least rectangle containing it inside (in the strict sense) and formed by these parallels (how exactly are these rectangles contructed? I assume that for each set $B$ in $U_1$ he takes the smallest rectangle that contains $B$. Is this correct?) Obviously the perimeter of the rectangle corresponding to an area of diameter $d$ is less than or equal to $8d$ (why not $4d$?)
In the general case the rectangles $u_{11},u_{12}, u_{13}, ..., u_{1k}$ corresponding to the areas of $U$ are overlapping. In this case we form a system of non-overlapping figures $v_{11},v_{12},\ldots,v_{1k}$ in the following way. We take $v_{11}=u_{11}$. Further, $v_{12}$ is the part of $u_{12}$ which is outside $u_{11}$; $v_{13}$ is the part of $u_{13}$ which is outside $u_{11}$ and $u_{12}$, and so on. It is clear that the sum of the perimeters of the figures $v_{1i}$ is less than the double sum for the rectangles $u_{1i}$...
Ok, if my assumption of how the rectangles are constructed is correct, this last part is clear to me. Now he splits the integral along $A_1$ into two parts: \begin{align*} \int_{A_1}f(z)dz=\sum'\int_{A_{1i}}f(z)dz+\sum''\int_{A_{1j}}f(z)dz, \end{align*} where the first summation $\sum'$ is extended over those squares $A_{1i}$ which are outside all figures $v_{1i}$, and $\sum''$ over the other squares. Then he claimes \begin{align*} \sum''\int_{A_{1i}}f(z)dz=\sum\int_{v_{1i}}f(z)dz. \end{align*} why? What if some parts of the figures $v_{1i}$ are outside of $A_1$? Next, he claimes that \begin{align*} \left|\sum'\int_{A_{i1}}f(z)dz\right|\geq c \end{align*} for some $c>0$ (he actually sais what $c$ is, but that's not my point). Why can this sum not be empty?
He then continues giving estimates of these sums and repeats the process constructing subsquares $A_2,A_3,A_3,...$ of $A_1$ converging to some point $z\in A_1$. He then claimes that $z$ does not belong to any of the systems $U_1,U_2,...$. why? What if $E$ is countable and dense in $D$? If we cover each point of $E$ by a disk, then the complement of this covering with respect to $A_1$ is empty...
As you can see, I totally do not understand the main idea of this proof. Any help is highly appreciated. Please let me know if I shall add more details to the question. Personal contact to me (for talking or access to the article) is also easily possible. Thank you in advance!