Showing that $E(X) = \int_{\Omega}X(\omega)d\mathbb{P}(\omega) = \int_\mathbb{R}xd\mathbb{P}_X(x)$

Question

Let $X$ be a real-valued random variable on the probability space $(\Omega, \mathscr{F}, \mathbb{P})$, so that X's expected value is $E(X) = \int_{\Omega}X(\omega)d\mathbb{P}(\omega)$. Suppose that $X$ has a law $\mathbb{P}_X$, so that $\mathbb{P}_X(A) = \mathbb{P}(X \in A)$. How could you then argue that $$E(X) = \int_\mathbb{R}xd\mathbb{P}_X(x).$$ The initial step is easy, as $$\int_\mathbb{R}xd\mathbb{P}_X(x) = \int_{x \in X[\Omega]}xd\mathbb{P}_X(x) = \int_{\Omega}X(\omega)d\mathbb{P}_X(X(\omega))$$ But I'm not entirely sure how to argue that $\mathbb{P}_X(X(\omega)) = \mathbb{P}(\omega)$. I've understood that the "no-brainer" approach is to show the claim for

1. for indicator functions,

2. non-negative simple functions,

3. non-negative measurable functions,

4. measurable functions

but I'd also like to understand the bigger picture of why we can swap the measures as we do.

Edit: In the event that someone still stumbles across this question, what I am really looking for is insight into the bigger picture, like what @William M. suggested with the pushforward measure, and the literacy recommendation related to this same problem.