Let $\nu$ and $\mu$ be two measures that are absolutely continuous w.r.t. some dominating measure $\lambda$, i.e. $\nu(dx) = \nu(x) \lambda(dx)$ and $\mu(dx) = \mu(x) \lambda(dx)$. Assume furthermore that $\mu$ dominates $\nu$. I am trying to show that $\frac{d\nu}{d\mu} (x) = \nu(x) / \mu(x).$
We have for any $C \in \mathcal{F}$, $$ \nu(C) = \int_C \nu(x) \lambda(dx) = \int_C \frac{d\nu}{d\mu}(x) \frac{d\mu}{d\lambda}(x) \lambda(dx)$$ and $\lambda$-a.e. we have that $\nu(x) = \frac{d\nu}{d\mu}(x) \mu(x)$. However i'm failing to conclude from here. Since the support of $\mu$ (which we denote by A) contains that of $\nu$ we could divide by $\mu(x)$ and set $\frac{d\nu}{d\mu}(x) = \frac{\nu(x)}{\mu(x)}$ on $A$ and set an arbitrary value on the complementary of $\bar{A}$. Since $\mu(\bar{A}) = 0$ i guess this is not an issue?
I am more generally struggling with the following. Assume that $\nu$ and $\mu$ are instead singular, i.e. there exists a set $B$ such that $\nu(B) = 0$ and $\mu(\bar{B}) = 0$. In measure theory the convention is $\infty \cdot 0 = 0$ so its seems that there is no issue with writing $$ \nu(\Omega) = 1 = \int \nu(x) \lambda(dx) = \int \frac{\nu(x)}{\mu(x)} \mu(x) \lambda (dx) = \int \frac{\nu(x)}{\mu(x)} \mu(dx) = \int_{\bar{B}} \frac{\nu(x)}{\mu(x)} \mu(dx) + \int_B \frac{\nu(x)}{\mu(x)} \mu(dx)$$ however this doesn't make sense since the term in the right most equality is $0$ using our convention. I guess my question is, what is stopping us from multiplying and then dividing by $\mu(x)$?