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$$ \lambda(S)=\sum_{\mathcal A\text{ atomic}} \bar\mu(S\cap A)\frac{\text{Area}(A\cap S\cap Z^c)}{\text{Area}(A\cap Z^c)}. $$

$$ \lambda(S)=\sum_{A\text{ atomic}} \bar\mu(S\cap A)\frac{\text{Area}(A\cap S\cap Z^c)}{\text{Area}(A\cap Z^c)}. $$

Yes. I agree with your interpretation of my answer. One way to understand a copula is as a probability measure $\mu$ on $[0,1]^2$ such that $\mu(A\times [0,1])=\mu_1(A)$ and $\mu([0,1]\times B)=\mu_2(B)$ for any subsets $A$ and $B$ of $[0,1]$ where $\mu_1$ and $\mu_2$ are the probability distributions on the first and second coordinates respectively. (This is also called a coupling of the two measures).

Yes. I agree with your interpretation of my answer. One way to understand a copula is as a probability measure $\mu$ on $[0,1]^2$ such that $\mu(A\times [0,1])=\mu_1(A)$ and $\mu([0,1]\times B)=\mu_2(B)$ for any subsets $A$ and $B$ of $[0,1]$ where $\mu_1$ and $\mu_2$ are the probability distributions on the first and second coordinates respectively. (This is also called a coupling of the two measures). To recover the copula from the measure, you define $C(a,b)=\mu([0,a]\times[0,b])$.

EDIT: More details added in response to Q2 added in OP

Yes. I agree with your interpretation of my answer. One way to understand a copula is as a probability measure $\mu$ on $[0,1]^2$ such that $\mu(A\times [0,1])=\mu_1(A)$ and $\mu([0,1]\times B)=\mu_2(B)$ for any subsets $A$ and $B$ of $[0,1]$ where $\mu_1$ and $\mu_2$ are the probability distributions on the first and second coordinates respectively. (This is also called a coupling of the two measures).

In this language a subcopula is a measure on $[0,1]^2$ which is only defined on a sub-algebra $\mathcal B_1\otimes \mathcal B_2$ where $\mathcal B_1$ is the $\sigma$-algebra generated by intervals with your endpoints $\mathcal S_1$ and $\mathcal B_2$ is the $\sigma$-algebra generated by intervals with intervals with your endpoints $\mathcal S_2$. Another way to say this (at least in the case of finite $\mathcal S_1$ and $\mathcal S_2$) is that this $\sigma$-algebra consists of all unions of atomic rectangles. A subcopula is then extended to a copula if there is a measure on $[0,1]^2$ that agrees with the subcopula measure on the $\sigma$-algebra $\mathcal B_1\otimes \mathcal B_2$.

If $\bar\mu$ is a copula extending the subcopula $\mu$; and if the union of the zero measure rectangles does not contain any atomic rectangle, let $Z$ denote the union of the zero measure rectangles. You can then define a new copula extension by

$$ \lambda(S)=\sum_{\mathcal A\text{ atomic}} \bar\mu(S\cap A)\frac{\text{Area}(A\cap S\cap Z^c)}{\text{Area}(A\cap Z^c)}. $$

The condition ensures that you are not dividing by zero. To see that $\lambda$ is a copula, it suffices to check that it agrees with $\bar\mu$ on the atomic rectangles. If $R$ is an atomic rectangle, the only contribution to the sum comes from $A=R$. and that term is $\bar\mu(R)\text{Area}(R\cap Z^c)/\text{Area}(R\cap Z^c)$, which is equal to $\bar\mu(R)$.

To see that no mass is put on $Z$, if you substitute $S=Z$, then the numerator of the fraction is zero, so that $\lambda(Z)=0$. This finishes the proof.

There is nothing about this proof that is two-dimensional, so as long as the union of your zero-measure boxes does not contain any atomic boxes, this argument is fine.