Is a rational of two measures a measure?

Question

Let $\mu,\nu\in \mathcal{M}(\mathbb{R}^2)$ be Radon measures and $\sigma(\mathbb{R}^2)$ the Borel sets of $\mathbb{R}^2$. I want to show properties for the map $$ \gamma: \sigma(\mathbb{R}^2)\to \mathbb{R},\quad \Omega\mapsto\gamma(\Omega)=\begin{cases}\frac{\mu(\Omega)}{\nu(\Omega)} & \nu(\Omega)\neq0 \\ 0 & \nu(\Omega)=0\end{cases}. $$ One question that I have is: "Is $\gamma$ even a finite measure?"

My attempt:

If $\nu\ll\mu$, the according to Radon-Nikodym there must be an $f$ such that $$ \nu(\Omega) = \int_\Omega f(\omega)d\mu(\omega) $$ for all $\Omega\in\sigma(\mathbb{R}^2)$. If $f$ could depend on $\Omega$, then $$ f(\Omega) = \begin{cases}(\nu(\Omega))^{-1} & \nu(\Omega) \neq 0 \\ 0 & \nu(\Omega) = 0\end{cases}. $$ However, $f$ cannot.

On the other hand if we have two measure $\mu_1,\mu_2\in \mathcal{M}(\mathbb{R}^2)$, then we can define a product measure $$ \mu_{1,2}(\Omega_1\times\Omega_2):=(\mu_{1}\otimes\mu_2)(\Omega_1\times\Omega_2)=\mu_{1}(\Omega_1)\mu_{2}(\Omega_2) $$ for all $\Omega_1,\Omega_2\in\sigma(\mathbb{R}^2)$. So it seems that I could write $\gamma$ as $$ \gamma(\Omega)= (\mu\otimes\nu_{-1})(\Omega\otimes\Omega) = \mu(\Omega)\nu_{-1}(\Omega) $$ with $\nu_{-1}$ given by $$ \nu_{-1}(\Omega) = \begin{cases}(\nu(\Omega))^{-1} & \nu(\Omega) \neq 0 \\ 0 & \nu(\Omega) = 0\end{cases}. $$ This seems to work out, when we can make sure the $\nu_{-1}(\Omega)$ does not blow up. So, consider a sequence of sets $(\Omega_n)_n$ such that $\Omega_n\in\sigma(\mathbb{R}^2)$, $|\mu|(\Omega_n)\neq 0$ and $$ \lim_{n\to\infty}|\nu(\Omega_n)| = 0. $$ Am I correct is stating that $\nu_{-1}$ is a finite measure (and by extension $\gamma$), since $$ |\nu_{-1}|(\Omega_n) < \infty $$ for all $n$?

NB: I phrased it as if $\nu$ is a Radon measure, but I am most interested in the case that $\nu$ is the Lebesgue measure.

Answer 1

No. When adding fractions, you don't expect $$ \frac{a+b}{c+d} =?\ \frac{a}{c} + \frac{b}{d} $$ which would have to hold for the quotient of two measures to be a measure.

If you really care about those quotients, see Functions resembling quotients of measures. Your function first appears at the paragraph labelled 1.9.

Answer 2

When $\mu$ and $\nu$ are arbitrary measures, the answer is no. To see that take two measures $\mu,\nu\in\mathcal{M}(\mathbb{R}^2)$ and a set $B\in\sigma(\mathbb{R}^2)$ such that $\nu(B)=0$. Observe that $$ 0 = \gamma(A\cup B)-\gamma(A)-\gamma(B) \\ = \frac{\mu(A)+\mu(B)}{\nu(A)} - \frac{\mu(A)}{\nu(A)} $$ for all $A\in\sigma(\mathbb{R}^2)$. This implies that $\mu(B)=0$ must hold, and thus $\mu$ cannot be an arbitrary measure. It must be at least absolutely continuous with respect to $\nu$.