I am reading these lecture notes, section 2.3 (pg 4), and I've become very confused about relationship between
The induced measure $\mu_X$ -- a random variable $X: \Omega \rightarrow S$ with the original measure $\mu$ induces $\mu_X(B) = \mu(X^{-1}(B))$.
The measure $\nu(A) = \int_A f\ d\mu$ corresponding to a density function $f:\Omega \rightarrow \mathbb{R}^{0+}$.
The notes compare the definition of density function I'm familiar with: $Pr(X \leq a) = \int_{-\infty}^a f(x)\ dx$ with the measure-theoretic equivalent: $\mu_X(B) = \int_B f\ d\lambda$ where $\lambda$ is the Lebesgue measure.
I'm trying to reconcile the original definition of $\mu_X$ with the new one, and I can't see why it should be the case that $\mu_X(B) = \int_B f\ d\lambda = \int_{X^{-1}(B)}1\ d\mu$ where the RHS is just another way of writing the original $\mu(X^{-1}(B))$.
I'm also confused because $\nu$ and $\mu_X$ are written so similarly you'd suspect they're the same thing, but $\mu_X$ is a measure on $(S, \mathcal{A})$ whereas $\nu$ is a measure on $(\Omega, \mathcal{F})$, so this is clearly not possible. But then I'm not sure what was the point of $\nu$.
I think this is somehow related to one of the theorems stated in the text: $\int g\ d\nu = \int f g\ d \mu$ (when $f, \nu, \mu$ are related as described above), but if I use this by plugging in $\lambda \rightarrow \mu, I_B \rightarrow g $ ($I_B$ being the indicator function for set $B$)) I end up with $\int I_B d\nu$ where $\nu(A) = \int_A1\ d\lambda$, which still doesn't get me anything I want.
What am I missing?
For any random variable $X$ we always have a measure $\mu_X$ on the Borel sigma algebra of the real line defined by $\mu_X(B)=P(X^{-1}(B))$. In general there is no density function of $f$. We say that $X$ has a density if there exist a non-negative measurable function $f$ such that $P(X^{-1}(B)=\int_B f(x)dx$ for all Borel sets $B$. In 2) $\mu$ is not $\mu_X$ but it is the Lebesgue measure. $\nu$ is same as $\mu_X$ and we have $\mu_X(B)=\int_B f(x)dx$ for all Borel sets $B$.