I am lost in the measure theoretic definition of a probability distribution. I am a computer science master student and although I have some mathematical background, I never had a formal training in measure theory.
Let $(\Omega, \mathcal{F},\mathbb{P})$ be a probability space and $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ be a mesurable space. Let $X:\Omega \rightarrow \mathbb{R}$ be a random variable on $(\mathbb{R}, \mathcal{B}(\mathbb{R})).$
The probability distribution of $X$ is the pushforward measure
(1) $\mathbb{P}_X (B) = \mathbb{P}(X^{-1}(B)) = \mathbb{P}(\{\omega\in \Omega \mid X(\omega) \in B\}), ~~~~ \forall B\in \mathcal{B}(\mathbb{R}).$
Can we say without any loss of generality that $\mathbb{P}_X(B) = \int_{x\in B} \mathbb{P}(dx)$ such that
(2) if $X$ is discrete then $\mathbb{P}_X(B) = \sum_{x\in B} \mathbb{P}(X=x);$
(3) and if $X$ is continuous then $\mathbb{P}_X(B) = \int_{x\in B} f(x) dx,$ where $f$ is a density function.
Question 1: Is it true?
Question 2: How can we go from (1) to (2)? What are the theorems to invoke?
Question 3: Can we justify (1) to (3) with the Radom-Nikodym theorem?
Question 4: I have trouble finding $\mathbb{P}_X(B) = \int_{x\in B} d\mathbb{P}(dx)$ written anywhere. Is it wrong? Or maybe trivial?
Thank you very much!
Note that we have
$$\mathbb{P}_X(A) = \mathbb{P}(X \in A) = \int_{\{X \in A\}} d\mathbb{P}$$
so (4) is definitely true.
If $X$ is absolutely continuous, then by definition there is a density function $g$ such that $$\mathbb{P}_X(A) = \int_A g d \lambda$$
You don't need Radon-Nikodym. It's the definition.
For (2),
$$\mathbb{P}_X(B) = \mathbb{P}_X\left(\bigcup_{x \in B} \{x\}\right) = \sum_{x \in B} \mathbb{P}_X(\{x\}) = \sum_{x \in B} \mathbb{P}(X = x)$$
note that the union becomes a sum because $X$ is discrete and we can view this as a countable sum (to be really formal, we should sum over all elements with non-zero probability).