Expectation and sign of the Radon-Nikodym derivative I am new here. I have some questions about the Radon-Nikodym derivative. I hope someone is willing to help me with these. The questions are stated below. Also I added my attempts to the problem to below all questions.

Question 1)

Let $(\Omega, 2^{\Omega},P)$ be a probability space such that $\Omega$ has a finite number of elements, and let $P(\omega)&gt;0, \forall \omega \in \Omega$. Let $Q$ be a probability measure, equivalent to $P$. Prove that there is a random variable $Z$ s.t. $Q(F)=E^{P}(1_{F}(\omega)Z), \forall F\in 2^{\Omega}$

Now it turns out that this random variable $Z$, denoted as $Z=\frac{dQ}{dP}$ is called the Radon-Nikodym derivative. 

Question 2)

Prove that this random variable is nonnegative and has expectation 1 w.r.t. the probability measure $P$. Hint (to prove that this R.V. is non-negative almost sure): you could use $(-\infty,0)=\cup_{n \geq1, n \in \mathbb{N}}(-\infty,-\frac{1}{n}) $ 



My attempts:

Q1] $E^{P}(1_{F}Z)=\sum_{\omega \in \Omega}1_{F}(\omega)Z(\omega)P(\omega)=\sum_{\omega \in F}Z(\omega)P(\omega)=Q(F)=\sum_{\omega \in F}Q(\omega)$.

Hence $Z(\omega)=\frac{Q(\omega)}{P(\omega)}, \forall \omega \in F$, $\forall F \in 2^{\Omega}$. is a random variable s.t. $Q(F)=E^{P}(1_{F}(\omega)Z), \forall F\in 2^{\Omega}$.

Q2] I think I am able to prove that this random variable $Z$ has expectation 1. I would do this is follows, however I am not sure!
$E^{P}(Z)=E^{P}\frac{Q(\omega)}{P(\omega)}=\sum_{\omega \in \Omega} P(\omega)\frac{Q(\omega)}{P(\omega)} = \sum_{\omega \in \Omega}Q(\omega)=1 $

So actually my biggest problem is to prove that this random variable $Z$ is nonnegative almost sure. I really did think about it, but can't come up with a reasonable answer where I also use the given hint. I Hope you are willing to help me with this, since I have shown now that I did think about these questions. However, I'm really stuck with this last problem, that $Z$ is nonnegative almost sure.

Yes indeed, $Q(\omega)$ and $P(\omega)$ are bigger than zero by definition and therefore the ratio is bigger than zero. But My problem still is that I did not use the hint which is given which feels wrong to me. Also, I think that $\frac{Q(\omega)}{P(\omega)}$ is just an example of such a $Z$. So in that case, I've proven only for this special case that the sign is positive almost sure and that it's expectation is one. Or am I wrong with this?

Thanx in advance!
I think I might have found the answer for the nonnegativity.

Let $F_n = Z^{-1}[(−∞,−(1/n))] = \{\omega \in \Omega : Z(\omega) \in (-\infty,-(1/n))\}$.
Note: $F_n \in \mathcal{F}$, since $Z$ is a random variable. 
Suppose $F_n \neq \emptyset$. Then we get:

$Q(F) = E^P[Z\chi_F]&lt;0$. This contradicts with the fact that Q is a probability measure.
Hence, $F_n = \emptyset$. Now we find:

$Z^{-1}[(-\infty,0)] = Z^{-1}[\cup_{n\in\mathbb{N}}(-\infty,(1/n))] = \cup_{n\in\mathbb{N}}Z^{-1}[(-\infty,(1/n))] = \cup_{n\in\mathbb{N}}F_n = \cup_{n\in\mathbb{N}}\emptyset = \emptyset $.

Hence, there exists no $\omega \in \Omega$ such that $Z(\omega)&lt;0$.

I always mess up with almost sure things, so maybe you can find how to include that, but I think this is most of the proof.
I think I might have found the answer for the nonnegativity.

Let $F_n = Z^{-1}[(−∞,−(1/n))] = \{\omega \in \Omega : Z(\omega) \in (-\infty,-(1/n))\}$.
Note: $F_n \in \mathcal{F}$, since $Z$ is a random variable. 
Suppose $F_n \neq \emptyset$. Then we get:

$Q(F) = E^P[Z\chi_F]&lt;0$. This contradicts with the fact that Q is a probability measure.
Hence, $F_n = \emptyset$. Now we find:

$Z^{-1}[(-\infty,0)] = Z^{-1}[\cup_{n\in\mathbb{N}}(-\infty,(1/n))] = \cup_{n\in\mathbb{N}}Z^{-1}[(-\infty,(1/n))] = \cup_{n\in\mathbb{N}}F_n = \cup_{n\in\mathbb{N}}\emptyset = \emptyset $.

Hence, there exists no $\omega \in \Omega$ such that $Z(\omega)&lt;0$.

I always mess up with almost sure things, so maybe you can find how to include that, but I think this is most of the proof.