let $p(dx)$ and $q(dx)$ two densities of representing two $L^1(\mathbb{P})$ random variables: $P$ and $Q$
if $p(x) = 0$ $=>$ $q(x)=0$ then is $P=Q$ in law ?
my intuition is based on Radon Nickodym's theorem wich will assure existance of a function $f$ s.t. $\forall B \in \mathcal{B}(\mathbb{R})$
$\int_B q(x)dx = \int_B f(x)p(x)dx$
but then this would imply $ f \equiv 1$ since q is a density and so $q = p$ a.s. (by a density argument or whatever)?
On
The fact that the two distributions are equal equivalent (or one is absolutely continuous wrt the other) does not tell you anything about the random variables. You could just take two variables with the same density which are not a.s. equal.
What maybe you were looking for is the following result.
Take a probability space $(\Omega, \mathcal{F}, \mathbb{P})$, with some random variable $X$ defined on it. Assume that the law of $X$ under $\mathbb{P}$ is $\mu$. If you have another distribution $\nu << \mu$ such that $\nu(dx) = \mu(dx) \lambda(x)$, then you can find a probability measure $\mathbb{Q}$ on $\Omega$ such that the law of $X$ under $\mathbb{Q}$ is $\nu$. This probability measure $\mathbb{Q}$ is given by $$ \mathbb{Q}(A) = \mathbb{E}^{\mathbb{P}} \mathbb{I}_A\lambda(X) $$
Consider $P \sim \mathcal{N}(0,1)$ and $Q \sim \mathcal{N}(0,2)$. Then obviously we have absolute continuity in both ways but we do not have $p(x)=q(x)$ $\lambda$-a.e. where $\lambda$ stands for the Lebesgue measure on $\mathbb{R}$.