How does one go about finding the Radon-Nikodym derivative? I've seen the following exercise.
Let $(\mathbb{N}, \mathcal{P}(\mathbb{N}), \mu)$ be a measure space, where $\mu$ is the counting measure. Let $\nu : \mathcal{P}(\mathbb{N}) \rightarrow [0, \infty]$ be defined as $\nu(A) = \sum_{i \in A} 2^{-i}$.
How does one compute $\frac{d\nu}{d\mu}$?
To show that $f = \frac{d\nu}{d\mu}$ it suffices to show that $$\int_A f d\mu = \nu(A) \tag{1}$$ for all $A \subset \mathbb{N}$.
Since $\nu(A) = \sum_{i \in A}2^{-i}$, it's easy to check that $f$ defined by $$f(i) = 2^{-i}, \ \ \forall i \in \mathbb{N}$$ satisfies (1).
Indeed, $$\int_A f d\mu = \sum_{i \in A}f(i)\mu(\{i\}) = \sum_{i \in A}2^{-i}(1) = \nu(A).$$
More generally, if $X = (X, \mathcal{P}(X))$ is a discrete measurable space, and $\nu$ is a measure on $X$, then $\nu \ll \mu$, with $\mu$ counting measure, and $f$ defined by
$$f(x) = \nu(\{x\}), \ \ \forall x \in X$$
is a version of $\frac{d\nu}{d\mu}$ because (as before)
$$\int_A fd \mu = \sum_{x \in A}f(x) \mu(\{x\}) = \sum_{x \in A}\nu(\{x\})(1) = \nu(A).$$
Outside of the discrete setting, it may be quite difficult to compute $\frac{d\nu}{d\mu}$.