Let $\mathcal{A}$ denote the collection of all subsets A of an uncountable set $\Omega$ for which either A or $A^c$ are countable. Let $\mu(A)$ denote the cardinality of A. Define $\phi(A)$ equal to $\emptyset$ or $\infty$ according as $A$ is countable or uncountable. Show that $\phi ≪ \mu$. Then show that the Radon-Nikodym theorem fails.
This is an assignment. I found that this problem is not rigorous unless in this case we treat $+\infty$ as a number; and this equation, $+\infty+\infty=+\infty$ holds where we do not distinguish the cardinality of all the rational number, $\aleph_0$, and the cardinality of real number, $\aleph_1$, and so on.
If we treat $\infty$ as a number, it is easy to show that $\mathcal{A}$ is a $\sigma$-field, $\mu$ and $\phi$ a measure and $\phi ≪ \mu.$
My question is: Does this make sense $\aleph_0+\aleph_1+\aleph_2$? (Set theory is not my suit.)