Prove that $D = ${$ A \in \mathscr{F}$ $: \mu (A)$ $= \nu(A)$ } is not a $\sigma$-algebra Problem
Consider the following set:

For measures $\mu , \nu$ on the measurable space $(\Omega, \mathscr{F})$ where $\mu (\Omega) = \nu (\Omega) < + \infty$, consider the set $D = ${$ A \in \mathscr{F} : \mu (A) = \nu(A)$ }. We want to determine if this is a $\sigma$-algebra.

This is a set that is often used as a standard example of a d-system (Dynkin system). I have found that this is not a $\sigma$ algebra, however, it is unclear to me how to construct a counterexample to show this fact.
Progress
Of course, given the fact that it is a d-system means that it will definitely contain $\Omega$ and will be closed under complements. Therefore, in order to find a counterexample, we must look for measures that ensure that the set $D$ is not closed under countable unions.
Looking online has produced a few different examples of possible counterexamples, however, as this was an exercise in a textbook I am working through, I am more interested in understanding how to construct a counterexample as opposed to just copying down a specific example that works in this instance without actually learning to construct this on my own.
Example Counterexample
The counterexample used in the textbook is the following:

Consider $D$ (as defined above) in the following setup:
Sample space: $\Omega = ${$1,2,3,4$}; Sigma Algebra: $\mathscr{F} = P(\Omega)$ (power set); Measures: $\mu = δ_{\{1\}} +δ_{\{2\}}$ and $\nu = δ_{\{3\}} +δ_{\{4\}}$.

If I didn't have access to the solutions, I never would have been able to construct this example on my own. What is the intuition behind something like this (or behind any possible answer for this question)? How can I learn to construct examples of my own here?
 A: My way of thinking about it after reading the first lines of your question.

It is enough to find a situation with $\mu(A)=\nu(A)$, $\mu(B)=\nu(B)$ and $\mu(A\cup B)\neq\nu(A\cup B)$.
A handsome partition of $\Omega$ that comes to mind is then the collection: $$\mathcal P=\{A\cap B,A\cap B^c,A^c\cap B,A^c\cap B^c\}$$ We need a $\sigma$-algebra that contains the elements of $\mathcal P$ and keeping things small and simple we take the collection of unions of elements of $\mathcal P$. Every measure on this $\sigma$-algebra is completely determined by the values that it takes on the elements of $\mathcal P$.
Now is it possible to find values for $\mu$ and $\nu$ on these sets in such a way that the conditions mentioned above are satisfied?
Yes, it is.
As an example let $\mu(P)=2$ for every $P\in\mathcal P$ and let $\nu(A\cap B)=\nu(A^c\cap B^c)=1$ and $\nu(A\cap B^c)=\nu(A^c\cap B)=3$.
Then $\mu(A)=\nu(A)=4$, $\mu(B)=\nu(B)=4$ and $\mu(A\cup B)=6\neq7=\nu(A\cup B)$.
Also we have $\mu(\Omega)=\nu(\Omega)=8<+\infty$.
The Dynkin-system that arises from this is:$$\mathcal D=\{\varnothing, A, B, A^c,B^c,\Omega\}$$ where the sets in $\mathcal P=\{A\cap B,A\cap B^c,A^c\cap B,A^c\cap B^c\}$ are not empty.
