Density of a Probability measure with respect to an equivalent one Let $\mathbb{P_1}$ and $\mathbb{P_2}$ be two equivalent measures and $Z=d\mathbb{P_1}/d\mathbb{P_2}$ the density of $\mathbb{P_1}$ with respect to $\mathbb{P_2}$. Prove: 
1) Uniqueness of Z up to $\mathbb{P_2}$-null sets,
2) $Z>0$ $\mathbb{P_2}$ almost surely,
3) $X \in L^{1}(\mathbb{P_1})$ if and only if $XZ \in L^{1}(\mathbb{P_2})$ and in this case $\mathbb{E_1}(X) = \mathbb{E_2}(XZ)$
4) $1/Z$ is the density of $\mathbb{P_2}$ with respect to $\mathbb{P_1}$
My approach for part 1) would be: let $Z_1$ and $Z_2$ two of those, then $Z_1=Z_2$ leads to $d\mathbb{P_1}/d\mathbb{P_2}=d\mathbb{P_1}/d\mathbb{P_2}$ and $d\mathbb{P_1}d\mathbb{P_2}=d\mathbb{P_1}d\mathbb{P_2}$ but I don't know how to interpret this expression.  I think it has something to do with the radon nykodim derivative. Can you help me to understand this problem?
 A: I am going to switch $\mathbb P_1$ and $\mathbb P_2$ to $\mu$ and $\nu$ for convenience.
By hypothesis, $\mu$ and $\nu$ are mutually absolutely continuous positive, finite measures. 
The R-N theorem gives us that $\mu(E)=\int_EZd\nu,\ $ where $Z$ is as in the question.  Now $Z\ge 0$ because the measures are positive, and if $Z_1$ satisfies the same condition as $Z$, then for all $\nu$-measurable sets $E$, we have $\int_EZd\nu=\int_EZ_1d\nu\Rightarrow \int_E |Z-Z_1|d\nu=0\Rightarrow Z=Z_1$ a.e.$\ \nu$.
Suppose $Z=0$ on a set $E$ of positive $\nu$-measure. Then, $\mu(E)=\int_EZd\nu=0$. On the other hand, $\nu\ll \mu$ so $\nu(E)=0$, which is a contradiction. Thus, $Z>0\ $ a.e. $\nu$.
Now, $d\mu=Zd\nu\Rightarrow \int_EXd\mu=\int_EXZd\nu$ for any $X\in L^1(\mu)$. (To see this, verify the formula for characteristic functions, then it will be true for simple functions and finally, apply the MCT to conclude). This shows that $XZ\in L^1(\nu)\Leftrightarrow X\in L^1(\mu).$ 
Finally, we have a R-N derivative $W$ for $\nu:\ \nu(E)=\int_EWd\mu.$ But also, by the remarks in the previous paragraph, $\int_E\frac{1}{Z}d\mu=\int_E\frac{1}{Z}\cdot Zd\nu\Rightarrow \nu(E)=\int_E\frac{1}{Z}d\mu$ and the uniqueness result of the first paragraph implies that $W=1/Z$ a.e.$\mu$. Note: $1/Z\in L^1(\mu)$ because $1/Z>0$ a.e. $\mu$ (use the same argument as that for $Z$.)
